Speed–time graphs questions with answers

Questions

1. Question 1Read the speed at a time

   The diagram shows the speed–time graph of a car over 95 seconds. Find the speed, in m/s, of the car when $t = 65$.

   Show the worked solution

   1. Step 1: Go up from the given time on the time axis to the graph, then across to the speed axisWhen $t = 65$, the graph is at $30$ m/s.
2. Question 2Acceleration and deceleration (the gradient)

   The diagram shows the speed–time graph of a train over 70 seconds. Find the deceleration, in m/s², of the train in the last 30 seconds.

   Show the worked solution

   1. Step 1: Fall in speed (the drop)$\text{Fall in speed} = 36 - 0$$\phantom{\text{Fall in speed}} = 36 \text{ m/s}$
   2. Step 2: Time taken (the run)$\text{Time} = 70 - 40$$\phantom{\text{Time}} = 30 \text{ s}$
   3. Step 3: Deceleration = fall in speed ÷ time (a deceleration is given as a positive number)$\text{Deceleration} = \dfrac{36}{30}$$\phantom{\text{Deceleration}} = 1.2 \text{ m/s}^2$
3. Question 3Distance = area under the graph

   The diagram shows the speed–time graph of a car over 50 seconds. Find the distance, in metres, travelled by the car in the first 30 seconds.

   Show the worked solution

   1. Step 1: Area A$\text{A}: 0 \text{ to } 30 \text{ s}, \text{ rectangle } 30 \times 12 = 360$
4. Question 4Total distance and average speed

   The diagram shows the speed–time graph of a car over 50 seconds. Find the average speed, in m/s, of the car for the 50 seconds.

   Show the worked solution

   1. Step 1: Area A$\text{A}: 0 \text{ to } 40 \text{ s}, \text{ triangle } \dfrac{1}{2} \times 40 \times 32 = 640$
   2. Step 2: Area B$\text{B}: 40 \text{ to } 50 \text{ s}, \text{ rectangle } 10 \times 32 = 320$
   3. Step 3: Distance = the total area$\text{Distance} = 640 + 320$$\phantom{\text{Distance}} = 960 \text{ m}$
   4. Step 4: Average speed = total distance ÷ total time$\text{Average speed} = \dfrac{960}{50}$$\phantom{\text{Average speed}} = 19.2 \text{ m/s}$
5. Question 5Find T or V from the distance

   The diagram shows the speed–time graph of a lorry that starts from rest and comes to rest again. The total distance travelled is 300 m. Find the value of $V$.

   Show the worked solution

   1. Step 1: Write its area in terms of V, equal to the distance$\dfrac{1}{2} \times 50 \times V = 300$
   2. Step 2: Simplify the left side$25 V = 300$
   3. Step 3: Divide both sides$V = \dfrac{300}{25}$$\phantom{V} = 12$
6. Question 6The speed part-way through a stage

   The diagram shows the speed–time graph of a car. Find the speed, in m/s, of the car when $t = 3$.

   Show the worked solution

   1. Step 1: Acceleration of this stage = change in speed ÷ time$\text{Acceleration} = \dfrac{9 - 0}{4}$$\phantom{\text{Acceleration}} = \dfrac{9}{4}$$\phantom{\text{Acceleration}} = 2.25 \text{ m/s}^2$
   2. Step 2: Time since this stage began$3 - 0 = 3 \text{ s}$
   3. Step 3: Speed = starting speed + acceleration × time$\text{Speed} = 0 + 2.25 \times 3$$\phantom{\text{Speed}} = 6.75 \text{ m/s}$
7. Question 7A speed given in km/h

   The diagram shows the speed–time graph of a train that starts from rest. The maximum speed, $V$, of the train is 72 km/h. Find the total distance, in metres, travelled by the train.

   Show the worked solution

   1. Step 1: Change V to m/s$V = \dfrac{72 \times 1000}{3600}$$\phantom{V} = 20 \text{ m/s}$
   2. Step 2: Distance = its area$\text{Distance} = \dfrac{1}{2} \times (20 + 55) \times 20$$\phantom{\text{Distance}} = 750 \text{ m}$
8. Question 8Two vehicles on one graph

   A car starts from rest and accelerates uniformly to a speed of $25$ m/s in 10 s, then travels at that speed. As it starts, a lorry travelling at a constant speed of 20 m/s passes it, in the same direction. The diagram shows their speed–time graphs. The car catches up with the lorry after $T$ seconds. Find the value of $T$.

   Show the worked solution

   1. Step 1: Distance of the car after T s: the area under its graph (a triangle, then a rectangle)$\dfrac{1}{2} \times 10 \times 25 + 25(T - 10)$$= 25T - 125$
   2. Step 2: Distance of the lorry after T s: a rectangle$20T$
   3. Step 3: The car catches up when the two distances are equal$25T - 125 = 20T$$5T = 125$
   4. Step 4: Solve for T$T = 25$