Forces and Motion 2.3 Acceleration formula, problem solving calculations and interpreting velocity-time graphs

Doc Brown's Physics exam study revision notes

Q1 A car accelerates from 10 m/s to 30 m/s in 15 seconds.

Calculate the acceleration of the car.

a (m/s²) = Δv (m/s) ÷ Δt (s) ÷ a

If u and v are the initial and final velocities

a = (v - u) / t = (30 - 10) / 15 = 20 / 15 = 1.3 m/s² (2 s.f.)

Q2 A train accelerates at 0.50 m/s² for 30 seconds.

What is the increase in speed of the train?

a = Δv ÷ t   so   Δv = a x t

increase in speed = 0.5 x 30 = 15 m/s (2 s.f.)

Q3 If a car accelerates at 0.30 m/s² from a standing start, how long will it take to attain a speed of 21 m/s?

a = Δv ÷ Δt   so   Δt = Δv ÷ a = (v - u) ÷ a, u and v are the initial and final velocities

time taken = (21 - 0) ÷ 0.30 = 21 s  (2 s.f.)

Q4 A car brakes sharply from moving at 30 m/s to 10 m/s in 4.0 seconds.

Calculate the deceleration of he car.

a = Δv ÷ Δt

If u and v are the initial and final velocities

a = (10 - 30) / 4.0 = 20 / 4 = -5.0 m/s²  (2 s.f.)

Note the minus sign for the acceleration - because the car is slowing down!

You can then say the deceleration is 5.0 m/s²  (2 s.f.)

Q5 The graph below summarises the first 100 seconds of a cyclist's race.

When calculating displacement (distance travelled), with this velocity - time graph, both triangles and rectangles are involved.

(a) (i) What is the initial uniform acceleration of the cyclist in this time period?

The initial acceleration is from 0 to 40 seconds.

speed = total change in speed / time taken = (10 - 0) / (40 - 0) = 10 / 40 = 0.25 m/s² (2 s.f.)

(ii) How far has the cyclist travelled in this period?

From a speed/velocity-time graph you calculate distance covered from the area under the graph for the time period involved.

Area under the graph from 0 to 40 seconds = 10 x 40 / 2 = 200 m  (its easy if the graph is linear)

(b) At which point is the cyclist's acceleration the greatest? and calculate the acceleration at this point.

The steeper the gradients the greater the acceleration.

The gradient is steepest from 90 to 100 seconds, so that is the time period of greatest acceleration.

a = ∆v / ∆t = (25 - 20) / (100 - 90) = 5 / 10 = 0.5 m/s²

Q6 The graph below summarises a 100 minute commuter train journey, but with some signal delays!

Watch out for min ==> hour conversions!

(a) At what time or times is the train moving at constant speed? Explain your reasoning.

From the 20th to 40th minute AND from the 50th to 70th minute.

The graph is horizontal, indicating constant speed (of 100 km/hour and 200 km/hour).

(b) Where is the greatest acceleration and what is its value in km/hour²?

Steepest gradient is from the 40th to the 50th minute.

Change in velocity = 200 - 100 = 100 km/hour

Time involved = 50 - 40 = 10 mins, 10/60 = 0.167 hours

gradient = 100 / 0.167 = 100 / 0.167 = ~600 km/hour²

(c) How far does the train travel in the first 20 minutes?

time = 20 / 60 = 0.333 hours, speed change = 100 - 0 = 100 km/hour.

area under graph = 100 x 0.333 / 2 = 16.7 km, ~17 km

Q7  A car travelling at 36 m/s skids off the road and hits a wall.

 If the car is brought to a halt in 2.0 seconds, what is the average deceleration of the car?

final velocity v = 0, initial velocity u = 36 m/s

acceleration a = (v - u) ÷ t = Δv ÷ t = (0 - 36) / 2.0 = -18 m/s²

So the average deceleration is 18 m/s²

Note: The acceleration due to gravity is 9.8 m/s², so in this crash you body would experience a force of nearly '2G', not good for your body!

Q8 The graph below profiles the speed of a cyclist at the start of a race.

NOT a linear graph!

You get the distance travelled by calculating or measuring off the graph, the area under the graph for the specified time interval. To help you appreciate the method, I've highlighted two areas in yellow to be estimated to answer the two questions below.

Step 1 is to work out the distance relating to one small square.

From the 'bold' graph lines larger square you have 5 x 5 = 25 small squares = 2 m/s x 20 s = 40 m,

so each small square = 40 / 25 = 0.625 m, so from this you can estimate the areas and distances.

(a) How far does the cyclist travel in the first 20 seconds of the race? (give your answers to 2 s.f.)

0-10s ~ 2 squares, 10-20s ~9, total ~11 squares = ~11 x 0.625 = 6.88, distance travelled ~6.9 m

(b) How far does the cyclist travel between the 30th and 40th second of the race.

total small squares  = 25 + 22 (25-~3) + 5.5 = ~52.5, distance travelled = 52.5 x 0.625 = ~32.8 = ~32 m

Do you agree with my estimations?

Q9 A van is travelling at 20 m/s has its brakes applied for 5.0 seconds.

If this produces a deceleration of -1.5 m/s², calculate the velocity of the van after braking.

a = Δv ÷ Δt

Δv = (v - u) = a x t  = -1.5 x 5.0 = -7.5

v - u = -7.5,  v - 20 = -7.5,  v = -7.5 + 20 = 12.5 m/s

 

Q10 If an object is fired vertically up in the air from ground level,

(a) Describe the energy store changes taking place and any assumptions you are making in your arguments. Assume the object leaves from, and returns to, the ground level.

When the object is thrown upwards it is given an initial maximum kinetic energy store.

As it rises the velocity decreases, its kinetic energy store (KE) decreases and its gravitational potential energy store (GPE) increases proportionately - this assume no other energy transfers are occurring.

This ignores the friction effect of air resistance - some KE is lost as thermal energy to increase the thermal energy store of the surroundings.

The decrease in velocity is due to the decelerating effect of the Earth's gravitational field.

When it reaches its maximum height, the object has its maximum GPE.

The object then accelerates back down to the ground, again due to the Earth's gravitational field.

At the same time, as the velocity increases, the object's GPE store is converted to its KE store and it hits the ground at its maximum velocity and with its maximum KE. This assumes no air resistance.

(b) Sketch the graph of velocity versus time for the whole flight.

There are two graph formats you can draw to illustrate the change in velocity with time and both are illustrated below.

Graph A is constructed in two halves.

From the maximum velocity x, the left half represents the decrease in velocity as the object's motion is slowed down by the Earth's gravitational field.

You can think of this as a negative acceleration i.e. deceleration - a negative gradient (a - m/s²).

The right half represents the increase in velocity as the object falls under gravity, attaining the maximum velocity x on hitting the ground.

You can think of this as a positive acceleration - a positive gradient (a + m/s²).

Graph A ignores the mathematical convention of a minus velocity on change of direction.

The bottom of the V, at time t, represents zero velocity as the object's motion is on the point of reversing direction to fall down.

Graph B is constructed as one continuous line.

This uses both positive and negative values of velocity, to represent the change in direction.

Therefore the velocity changes from +x to -x in a time of 2 x t, t being the point of zero velocity

 (c) Can you suggest a value for the gradients of the graph?

The gradients in each case would have the value of the Earth's gravitational acceleration/deceleration constant of 9.8 m/s².