Question 1: A ball is placed at rest at the top of a hill. It travels with constant acceleration for the first 12 second and reaches a speed of 4 m/s. It then decelerates at a constant rate of 0.1 m/s2 for 20 seconds. It then travels at a constant speed for a further 18 seconds. 


Draw a speed-time graph for the ball over the course of this 50 seconds.

Question 2: Below is a speed-time graph of a track cyclist during a race. Work out the total distance travelled by the cyclist over the course of the race.



In this second question, the AREA under the graph is the total distance travelled. The best way to approach these questions is to split the area, as shown, into smaller simple shapes, then work out the areas of each individually.


There are a few ways to do this, for example the triangles B and D could have joined the square C split vertically downwards to the 15 point on the “x” axis, making two trapeziums and the areas of these could be calculated using the formula for the area of a trapezium.


I chose to break the area as shown, and the calculations were:


Area of a Triangle:



Area of A = 75 

Area of B = 12.5 

Area of C = 100 

Area of D = 25 

Area of E = 600 


Total distance travelled is therefore the sum = 812.5m



Question 3: A driver makes the journey represented by the graph shown.


(a) Identify all of the sections of the journey where the driver is accelerating (not decelerating)


Sections A D E - where there is a positive gradient to the graph.


(b) When is the driver’s velocity constant?


All points where the graph line is horizontal, so sections B F and H


(c) When does the driver have the greatest acceleration?


At that point where the positive gradient is steepest, so section D


(d) In which two sections does the driver have the same deceleration?


At the points where the gradient (negative in each case) is the same, so sections C and G


Be aware that some questions will use “acceleration” to mean speeding up OR slowing down, as “deceleration” is also acceleration but is negative.



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