5.14 Acceleration and velocity-time graphs
10 exam questions worth 27 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
0 / 27
What is the unit of acceleration?
Choose one answer.
Correct Not quite: the answer is m/s²
Mark scheme
B: m/s²
Model answer
m/s² (B).
A motorbike accelerates from 5.0 m/s to 25 m/s in 4.0 s. Calculate its acceleration.
Mark scheme
a = (v − u) ÷ t (1); (25 − 5.0) ÷ 4.0 (1); 5.0 m/s² (1)
Model answer
a = (v − u) / t = (25 − 5.0) / 4.0 = 5.0 m/s².
A train slows from 30 m/s to rest in 50 s. Calculate its acceleration, and explain the sign of your answer.
Mark scheme
a = (0 − 30) ÷ 50 (1); −0.6 m/s² (1); negative because it is decelerating / the acceleration is in the opposite direction to the motion (1)
Model answer
a = (0 − 30) / 50 = −0.60 m/s². The negative sign shows the train is decelerating.
A car accelerates uniformly from rest at 3.0 m/s² over a distance of 150 m. Calculate its final speed.
Mark scheme
v² = u² + 2 a s = 0 + 2 × 3.0 × 150 (1); v² = 900 (1); 30 m/s (1)
Model answer
v² = 2 × 3.0 × 150 = 900, so v = 30 m/s.
A stone is dropped from the top of a cliff and takes 3.0 s to reach the bottom. Calculate its speed when it hits the bottom. (a = 9.8 m/s², ignore air resistance)
Mark scheme
v = u + a t, or a = (v − u) ÷ t (1); 0 + 9.8 × 3.0 (1); 29 m/s (1)
Model answer
v = a t = 9.8 × 3.0 = 29.4 ≈ 29 m/s.
The velocity-time graph for a cyclist is a horizontal line. State what this shows about the cyclist's acceleration and velocity.
Mark scheme
the acceleration is zero (1); the velocity is constant (1)
Model answer
The acceleration is zero and the velocity is constant.
Use the velocity-time graph in this lesson to calculate the total distance travelled by the object between 0 s and 10 s.
Mark scheme
area of the triangle = ½ × base × height (1); ½ × 10 × 20 (1); 100 m (1)
Model answer
Distance = ½ × 10 × 20 = 100 m.
A velocity-time graph for a car has a straight line sloping up from (0 s, 0 m/s) to (6 s, 18 m/s), then a horizontal line to (16 s, 18 m/s). Calculate the acceleration for the first 6 s, and the total distance travelled in 16 s.
Mark scheme
a = 18 ÷ 6 (1); 3.0 m/s² (1); triangle: ½ × 6 × 18 = 54 m (1); rectangle: 10 × 18 = 180 m (1); total 234 m (1)
Model answer
a = 18 / 6 = 3.0 m/s². Distance = triangle (½ × 6 × 18 = 54 m) + rectangle (10 × 18 = 180 m) = 234 m.
Describe the motion shown by a velocity-time graph that is a straight line sloping down from 20 m/s to 0 m/s over 10 s.
Mark scheme
the object is slowing down / decelerating (1); at a constant rate of 2 m/s² (1)
Model answer
The object decelerates at a constant rate of 2.0 m/s², coming to rest after 10 s.
A student says that, on a velocity-time graph, a horizontal line at 20 m/s means that the object is stationary. Explain why the student is wrong.
Mark scheme
the object is moving at a constant velocity of 20 m/s (1); a stationary object has a velocity of zero / would be on the horizontal axis (1)
Model answer
The velocity is 20 m/s, so it is moving at constant velocity. A stationary object would have a velocity of zero.
Found a mistake?