Linear Acceleration Revision Equations of Motion

1. 
   v = u + at
   
2. 
   s = ut + ½ at²
   
3. 
   v² = u² + 2as
   

v = final velocity

u = initial velocity

a = acceleration

s= displacement (not distance)

t = time
Solving problems using equations of motion.

1. 
   Write down v, u, a, s and t filling in the quantities you know
   
2. 
   Write down the three equations of motion.
   
3. 
   Decide which of the three equations has only one unknown in it.
   
4. 
   Substitute the known values in to this equation and solve to find the unknown.
   

If a graph is drawn of Velocity (y-axis) against Time (x-axis), the slope of the graph is the acceleration of the object.

Four points a, b, c and d lie on a straight level road.

A car, travelling with uniform retardation, passes point a with a speed of 30 m/s and passes point b with a speed of 20 m/s.

The distance from a to b is 100 m. The car comes to rest at d.

Find

1. 
   the uniform retardation of the car
   
2. 
   the time taken to travel from a to b
   
3. 
   the distance from b to d
   
4. 
   the speed of the car at c, where c is the midpoint of [bd].
   

1. 
   v² = u² + 2as
   

-500 = 200 a

a = - 2.5 m s^-2

2. 
   v = u + at
   

T = 4 s

3. 
   v² = u² + 2as
   

s = 80 m

4. 
   v² = u² + 2as
   

= 200

A car travels from p to q along a straight level road.

It starts from rest at p and accelerates uniformly for 5 seconds to a speed of 15 m/s.

It then moves at a constant speed of 15 m/s for 20 seconds.

Finally the car decelerates uniformly from 15 m/s to rest at q in 3 seconds.

1. 
   Draw a speed-time graph of the motion of the car from p to q.
   
2. 
   Find the uniform acceleration of the car.
   
3. 
   Find the uniform deceleration of the car.
   
4. 
   Find |pq|, the distance from p to q.
   
5. 
   Find the speed of the car when it is 13.5 metres from p.
   

2. 
   v = u + at
   

a = 3 m s^-2

3. 
   v = u + at
   

a = - 5

4. 
   distance = ½ (5)(15) + (20)(15) + ½ (3)(15)
   

= 360 m

5. 
   v² = u² + 2as
   

= 81 m

In the absence of air resistance, all objects near the Earth’s surface will fall with the same acceleration, acceleration due to gravity. The value of g on the surface of the Earth is 9.8 m s^-2.

• 
  Because we take the upward direction as positive, and because g is acting downwards, we take g to be minus (-) 9.8 m s^-2 when answering maths questions (i.e. the initial velocity is usually opposite in direction to acceleration).
  
• 
  If an object is released from rest it means that initial velocity is 0 (u = 0).
  
• 
  At the highest point of a trajectory, the (instantaneous) velocity is zero (object is not moving upwards or downwards).
  
• 
  Also at the highest point of the trajectory, while the velocity is zero, the object is still accelerating at -9.8 m s^-2.
  

Ball A is thrown up into the air and two seconds later ball B is thrown up. The balls collide after a further t seconds.

Key: For the collision, A is in the air for t seconds and B is in the air for (t-2) seconds

Or

B is the air for t seconds and A is in the air for (t+2) seconds.

The second option is preferable.

Two balls are thrown up and collide in the air

Key: s₁ = s₂

If a ball is thrown up vertically its velocity at the very top is zero.
Distance travelled

You need to be able to distinguish between the concepts of distance and displacement.

On the way up, distance and displacement will be the same, but not when the ball is coming down (distance is total distance travelled, while displacement is only the height above the ground).
Note that in our equations of motion s always corresponds to displacement, not distance, although in most cases these will be the same.

If you are asked for distance travelled you will first have to establish whether the ball is on the way up or the way down. For this you will have to calculate the time of collision and the time taken to reach max. height.

A particle falls from rest from a point P.

When it has fallen a distance 19·6 m a second particle is projected vertically downwards from P with initial velocity 39·2 m s⁻¹.

The particles collide at a distance d from P.

Find the value of d.



2011 (a)

A particle is released from rest at A and falls vertically passing two points B and C.

It reaches B after t seconds and takes seconds to fall from B to C, a distance of 2.45 m.

Find the value of t.

A particle is projected vertically upwards from the point p. At the same instant a second particle is let fall vertically from q.

The particles meet at r after 2 seconds.

The particles have equal speeds when they meet at r.

Prove that pr = 3rq .

2008 (a)

A ball is thrown vertically upwards with an initial velocity of 39.2 m/s.

1. 
   Find the time taken to reach the maximum height
   
2. 
   Find the distance travelled in 5 seconds.
   

A particle is projected vertically downwards from the top of a tower with speed u m/s. It takes the particle 4 seconds to reach the bottom of the tower.

During the third second of its motion the particle travels 29.9 metres.

Find

1. 
   the value of u
   
2. 
   the height of the tower.
   

A stone is thrown vertically upwards under gravity with a speed of u m/s from a point 30 metres above the horizontal ground. The stone hits the ground 5 seconds later.

1. 
   Find the value of u.
   
2. 
   Find the speed with which the stone hits the ground.
   

A particle is projected vertically upwards with an initial velocity of u m/s and another particle is projected vertically upwards from the same point and with the same initial velocity T seconds later.

1. 
   Show that the particles will meet () seconds from the instant of projection of the first particle
   
2. 
   Show that the particles will meet at a height of metres.
   

A ball is thrown vertically upwards with an initial velocity of 20 m/s.

One second later, another ball is thrown vertically upwards from the same point with an initial velocity of u m/s.

The balls collide after a further 2 seconds.

1. 
   Show that u = 17.75.
   
2. 
   Find the distance travelled by each ball before the collision, giving your answers correct to the nearest metre.
   

Here the acceleration is constant throughout and we are given information about different stages.

Usually we are given information on two sections of an objects travel; the first is from the beginning, and another section is straight after.

We need to get an equation for both and solve, but to do this the variables (particularly u) must represent the same number for both equations.

The only way to do this is to make the second equation represent the first two sections, i.e. bring it back to the beginning. This means the distance s must be the distance from the beginning.

2010 (b)

A particle passes P with speed 20 ms^-1 and moves in a straight line to Q with uniform acceleration.

In the first second of its motion after passing P it travels 25 m.

In the last 3 seconds of its motion before reaching Q it travels of PQ.

Find the distance from P to Q.

2003 (a)

The points p, q and r all lie in a straight line.

A train passes point p with speed u m/s.

The train is travelling with uniform retardation f m/s².

The train takes 10 seconds to travel from p to q and 15 seconds to travel from q to r, where | pq| = | qr | = 125 metres.

1. 
   Show that f =
   
2. 
   The train comes to rest s metres after passing r. Find s, giving your answer correct to the nearest metre.
   

A particle, with initial speed u, moves in a straight line with constant acceleration.

During the time interval from 0 to t, the particle travels a distance p.

During the time interval from t to 2t, the particle travels a distance q.

During the time interval from 2t to 3t, the particle travels a distance r.

Show that 2q = p + r.

Show that the particle travels a further distance 2r − q in the time interval from 3t to 4t.
2000 (a)

A stone projected vertically upwards with an initial speed of u m/s rises 70 m in the first t seconds and another 50 m in the next t seconds. Find the value of u.

Usually these are quite straightforward.

Where the objects are moving up a slope should use the line of slope as the x-axis, which means that you will have to calculate the component of gravity in that direction also.
2004 (b)

A car of mass 1200 kg tows a caravan of mass 900 kg first along a horizontal road with acceleration f and then up an incline α to the horizontal road at uniform speed.

The force exerted by the engine is 2700 N.

Friction and air resistance amount to 150 N on the car and 240 N on the caravan.

1. 
   Calculate the acceleration, f, of the car along the horizontal road.
   
2. 
   Calculate the value of α, to the nearest degree.
   

A mass of 8 kg falls freely from rest.

After 5 s the mass penetrates sand.

The sand offers a constant resistance and brings the mass to rest in 0.01 s.

1. 
   Find the constant resistance of the sand
   
2. 
   Find the distance the mass penetrates into the sand.
   

1. 
   Acceleration – constant velocity – deceleration
   

You need to be pretty nifty with algebra to solve these guys.

You will most likely have to make use of the fact that the area under each section = distance travelled.

Write out as many relevant equations as you can to begin with.

You will usually need to incorporate information from the acceleration/deceleration section below also.

2. 
   Acceleration / deceleration
   

Consider the diagram above

1. 
   By definition: acceleration a = v/t
   

Therefore a = v/t₁

Similarly b = v/t₂

2. 
   Acceleration a = tan
   

3. 
   Let’s assume that we are told in a question that the deceleration is twice the acceleration.
   

• 
  t₁ is twice t₂.
  
• 
  t₁ is two thirds of the total time T, t₂ is one third of the total time T.
  

• 
  Area 1 is twice Area 2 {assuming we are starting from rest and finishing at rest}.
  
• 
  Angle 2 = twice Angle 1 therefore β = 2α therefore tan β = 2 tan α
  
• 
  a = v/t (if starting from rest)
  
• 
  Final velocity v = at (if starting from rest)
  

A car accelerates uniformly from rest to a speed v in t₁ seconds.

It continues at this constant speed for t seconds and then decelerates uniformly to rest in t₂ seconds.

The average speed for the journey is .

1. 
   Draw a speed-time graph for the motion of the car.
   
2. 
   Find t₁ + t₂ in terms of t.
   
3. 
   If a speed limit of were to be applied, find in terms of t the least time the journey ^{would have taken, assuming the same acceleration and deceleration as in part (ii).}
   

A train accelerates uniformly from rest to a speed v m/s with uniform acceleration f m/s².

It then decelerates uniformly to rest with uniform retardation 2f m/s².

The total distance travelled is d metres.

1. 
   Draw a speed-time graph for the motion of the train.
   
2. 
   If the average speed of the train for the whole journey is , find the value of f.
   

A lift starts from rest. For the first part of its descent it travels with uniform acceleration f. It then travels with uniform retardation 3f and comes to rest.

The total distance travelled is d and the total time taken is t.

1. 
   Draw a speed-time graph for the motion.
   
2. 
   Find d in terms of f and t.
   

A train accelerates uniformly from rest to a speed v m/s.

It continues at this speed for a period of time and then decelerates uniformly to rest.

In travelling a total distance d metres the train accelerates through a distance pd metres and decelerates through a distance qd metres, where p < 1 and q < 1.

1. 
   Draw a speed-time graph for the motion of the train.
   
2. 
   If the average speed of the train for the whole journey is , find the value of b.
   

Points p and q lie in a straight line, where |pq| = 1200 metres.

Starting from rest at p, a train accelerates at 1 m/s² until it reaches the speed limit of 20 m/s.

It continues at this speed of 20 m/s and then decelerates at 2 m/s², coming to rest at q.

1. 
   Find the time it takes the train to go from p to q.
   
2. 
   Find the shortest time it takes the train to go from rest at p to rest at q if there is no speed limit, assuming that the acceleration and deceleration remain unchanged at 1 m/s² and 2 m/s², respectively.
   

A car, starting from rest and travelling from p to q on a straight level road, where pq = 10 000 m, reaches its maximum speed 25 m/s by constant acceleration in the first 500 m and continues at this maximum speed for the rest of the journey.

A second car, starting from rest and travelling from q to p, reaches the same maximum speed by constant acceleration in the first 250 m and continues at this maximum speed for the rest of the journey.

1. 
   If the two cars start at the same time, after how many seconds do the two cars meet?
   
2. 
   Find, also, the distance travelled by each car in that time.
   
3. 
   the start of one car is delayed so that they meet each other exactly halfway between p and q, find which car is delayed and by how many seconds.
   

If acceleration is constant throughout then use equations of motion rather than a diagram.

1. 
   ^{The car and the bus meet after t seconds. Find the value of t.}
   
2. 
   ^{Find the distance between the car and the bus after 48 seconds.}
   

A car is travelling at a uniform speed of 14 m s^-1 when the driver notices a traffic light turning red 98 m ahead.

Find the minimum constant deceleration required to stop the car at the traffic light,

1. 
   if the driver immediately applies the brake
   
2. 
   if the driver hesitates for 1 second before applying the brake.
   

Car A and car B travel in the same direction along a horizontal straight road.

Each car is travelling at a uniform speed of 20 m/s.

Car A is at a distance of d metres in front of car B.

At a certain instant car A starts to brake with a constant retardation of 6 m/s².

0.5 s later car B starts to brake with a constant retardation of 3 m/s².

1. 
   Find the distance travelled by car A before it comes to rest
   
2. 
   Find the minimum value of d for car B not to collide with car A.
   

Two particles P and Q, each having constant acceleration, are moving in the same direction along parallel lines. When P passes Q the speeds are 23 m/s and 5.5 m/s, respectively.

Two minutes later Q passes P, and Q is then moving at 65.5 m/s.

1. 
   Find the acceleration of P and the acceleration of Q
   
2. 
   Find the speed of P when Q overtakes it
   
3. 
   Find the distance P is ahead of Q when they are moving with equal speeds.
   

Two trains P and Q, each of length 79.5 m, moving in opposite directions along parallel lines, meet at o, when their speeds are 15 m/s and 10 m/s respectively.

The acceleration of P is 0.3 m/s² and the acceleration of Q is 0.2 m/s².

It takes the trains t seconds to pass each other.

1. 
   Find the distance travelled by each train in t seconds.
   
2. 
   Hence, or otherwise, calculate the value of t.
   
3. 
   How long does it take for 2/5 of the length of train Q to pass the point o?
   

A man runs at constant speed to catch a bus. At the instant the man is 40 metres from the bus, it begins to accelerate uniformly from rest away from him. The man just catches the bus 20 seconds later.

1. 
   Find the constant speed of the man.
   
2. 
   If the constant speed of the man had instead been 3 m/s, show that the closest he gets to the bus is 17.5 metres.
   

Use v = u + at for both particles. Then use v² = u² + 2as for both particles to get the required expression.

1. 
   Velocity-time graph
   
2. 
   You need to play around with lots of algebra. Get an expression for total time in terms of v and f, then use the fact that average speed = total distance/total time to get an answer of f = 1 m s^-2.
   

1. 
   Straightforward. Ans: t = 4 s.
   
2. 
   Need to distinguish between the concepts of distance and displacement.
   

Ans: total distance = 83.3 m

1. 
   You know v, u and t for Q (t is when Q passes P), so use this to work out a.
   

Now for P you have this same s, plus u and t, so use this to work out a for P.

Ans: a_Q = 0.5 m s^-2, a_P = 5/24 m s^-2.

2. 
   Use v = u + at to find v for P. Ans: v_P = 48 m s^-2.
   
3. 
   “When” they are moving at equal speeds  v_P = v_Q so get an expression for both and equate.
   

Ans: distance = 525 m

1. 
   It’s not obvious, but this is a ‘Train-track’ type question.
   

We can get an expression for h - the distance travelled in the first 2 seconds: we know u, a and t.

We can then do the same for the distance travelled in the first 3 seconds. S in this case is (h + 29.9) m.

Ans: u = 5.4 m s^-1

2. 
   Straightforward. Ans: s = 100 m.
   

1. 
   Straightforward.
   
2. 
   Tricky: One car – two accelerations
   

For each section write down the relationship between velocity, distance and time

To find the total time you need to find an expression for each of the 3 individual times (in terms of velocity and distance).

Then it’s just messy algebra to finish it out.

Ans: b = 1

2006 (a)

Acceleration / deceleration type question.

1. 
   Straightforward
   
2. 
   Messy algebra. Ans: d = 3/8 ft²
   

1. 
   Straightforward.
   
2. 
   Straightforward once you draw a diagram to help you verify that S_P + S_Q = 159. Then solve. Ans: t = 6 s.
   
3. 
   Straightforward. Ans: t = 3.1 s.
   

1. 
   Straightforward (once you note that the acceleration is minus). Ans: s = 33.3 m
   
2. 
   Note that there are two distances to take into account here. The first is to do with reaction-time distance, and the second is the normal stopping distance. Add these together but remember that you have to subtract the first distance of 33.3 m because the first car will have moved on by this distance. Ans: s = 43.3 m.
   

1. 
   Straightforward once you are familiar with the concept of F_net = ma.
   

Then it’s just Force_down – Force_up = ma, where Force_down = mg, and Force_up is due to resistance of the sand, which is what you are looking for.

2. 
   Straightforward. Ans: s = 0.245 m.
   

1. 
   Straightforward. s₁ = s₂. Note that first ball is in the air for 3 seconds and second ball is in the air for 2 seconds.
   

2. 
   You first have to establish for each ball whether it will be on the way up or the way down. Hint: look at their initial velocities and the time they taken to reach max. height. Draw a diagram for each ball to help you.
   

1. 
   Straightforward once you are familiar with F_net = ma.
   

2. 
   Similar to (i), except in this case the car and caravan are going uphill so you will have to and also resolve the weight into components along the plane and perpendicular to the plane, and proceed accordingly.
   

1. 
   Straightforward train-track question.
   
2. 
   Straightforward once you remember that all numbers must be in relation to point p, so total distance travelled before coming to rest is (250 + s). Remember also that a = -3.
   

1. 
   Man just catches bus, so at this time v_Man = v_bus. Also when the man catches up with the bus he will hae travelled 40 m more than the bus, so s_Man = (40 + s_Bus).You will need to play around with the various equations of motion and use lots of algebra to solve.
   

2. 
   Find the distance between them means find get an expression for the distance travelled by both and subtract one from the other. In this case you are asked to find the minimum distance, so anytime maximum or minimum is asked for it usually means you have to differentiate and let your answer equal 0 to find t (remember from maths how to find the maximum or minimum point on a curve? – this is one of the most important applications of differentiation).
   

1. 
   Straightforward. Note s = -30. Ans: u = 18.5 ms^-1.
   
2. 
   Straightforward. Ans: speed = 30.5 m s^-1 (remember strictly speaking ‘speed’ implies magnitude only, not direction, so we should ignore the minus sign).
   

1. 
   Train-track type question. Straightforward.
   
2. 
   Straightforward.
   

1. 
   Velocity-time graph. Lots of algebra. See notes for answering this type of question above.
   

2. 
   Acceleration-deceleration. Algebra. See notes above.
   

1. 
   Two balls are thrown up and collide in the air so remember that the key is s₁ = s₂. Remember also that if the second ball is in the air for t₁ seconds, then the first ball (which is obviously in the air for longer) is in the air for (T + t₁) seconds. Note also that you were asked for the time taken in relation to when the first particle was projected, so you may have to adjust your answer accordingly.
   
2. 
   Sub value for time into expression for s.
   

This involves a stone projected upwards, but is actually a type of train-track type question because all information must be with reference to the initial point of projection.

Ans: u = 56 m s^-1.

2000 (b)

1. 
   Velocity-time diagram
   

Ans: t = 215 s, s_p = 4875 m, s_q = 5125 m.

2. 
   Cars meet halfway  s_p = s_q.
   

A man runs after a bus and just catches it.

Key: v_Man = v_Bus. Why? It’s the word Just that’s crucial here. If the man was going quicker than the bus when he got up to it then you wouldn’t use the term “he just caught it”. On the other hand if the man was going slower than the bus then he wouldn’t catch up with it at all (at all).
Greatest gap

The greatest gap between them also occurs when v_man = v_Bus  (because if their speeds are unequal then the gap is either increasing or decreasing). Another way of solving this is getting an expression for the distance between them (s_Bus – s_man) and then differentiating and letting the answer = 0. i.e. d(s_Bus – s_man)/dt = 0.

‘Retardation’ is the scientific term for ‘deceleration’, i.e. acceleration is minus (strictly speaking we physicists would say that the car is simply accelerating in the ‘minus’ direction).