physics of movement and collisions

Unit 26 – Forensic Traffic Collision Investigation
Assignment 26.2 – Physics of Movement and Collision
Name- Muhammed Zain Gorji


Assignment 26.2 Task 2 -Physics of movement and collisions

Complete the worksheet that will demonstrate your competence using a range of physical
terms and relationships. Ensure you show all your working, including the formulae you use,
and include the appropriate units with your answers.

THINKING DISTANCE
1. A car is travelling along a straight road at 7.5 m s–1. How far will it travel in 35s?
v= (S/t)
S= (vt)
S= (7.5×35)
S= 262.5m

2. A car is travelling along a straight road at 8.5 m s–1. How far will it travel in 13mins?
v= (S/t)
S= (vt)
S= (8.5×780)
S= 6630m

3. A car is travelling along a straight road at 12.5 m s–1. How far will it travel in 1 hour
15mins?
v= (S/t)
S= (vt)
S= (12.5×4500)
S= 56250m
–1
4. A car is travelling along a straight road at 9.5 m s . The driver who is alert sees a
stationary car ahead and needs to stop. It takes the driver 2.5s to apply the brakes,
calculate the thinking distance.
v= (S/t)
S= (vt)
S= (9.5×2.5)
S= 23.75m

5. An intoxicated person is driving along a straight road at 9.5 m s–1. The driver sees a
stationary car ahead and needs to stop. It takes the driver 5.5s to apply the brakes,
calculate the thinking distance. Compare this distance with the one in Q4 and make an
appropriate comment about the dangers of driving whilst intoxicated.
v= (S/t)
S= (vt)
S= (9.5×5.5)
S= 52.25m

The thinking distance is 28.5m greater than the distance in Q4. This is because the driver
was intoxicated whilst driving and this causes increased confidence (more than what a

, Unit 26 – Forensic Traffic Collision Investigation
Assignment 26.2 – Physics of Movement and Collision
Name- Muhammed Zain Gorji


driver should be) and delayed reaction times, which together contribute to the risk of a
collision.

ACCELERATION AND BRAKING DISTANCE
6. A car is travelling along a straight road at 7.5 m s–1. It accelerates at 1.5 m s–2 for a
distance of 32 m. How fast is it then travelling?
v = U2+2(aS)
2

v2= 7.52+2(1.5×32)
v2= 152.25
v= 12.33896268
v= 12.34ms-1 to 2 decimal places

7. A car is travelling along a straight road at 5.0 m s–1. It accelerates at 2.5 m s–2 for a
distance of 25 m. How fast is it then travelling?
v2= U2+2(aS)
v2= 5.02+2(2.5×25)
v2= 150
v= 12.24744871
v= 12.25ms-1 to 2 decimal places

8. A car travelling at 15 m s–1 accelerates at 0.55 m s–2 for 25 s. Calculate the distance
travelled by the car in this time.
S= (Ut)+((At2)/2)
S= (15×25)+((0.55×252)/2)
S= 546.875m

9. A cyclist is travelling at 11 m s–1. She brakes so that she doesn’t collide with the wall
which is 15.5m away. Calculate the magnitude of her deceleration.
v2= U2+2(aS)
a= (v2-U2)/(2S)
a= (02-112)/(2×15.5)
a= 3.903225806
2
a= 3.90m/s to 2 decimal places

10. A cyclist is travelling at 10 m s–1. She brakes so that she doesn’t collide with the wall
which is 7m away. Calculate the magnitude of her deceleration.
v = U2+2(aS)
2

a= (v2-U2)/(2S)
a= (02-102)/(2×7)
a= 7.142857143
a= 7.14m/s2 to 2 decimal places

11. A car is travelling along a straight road at 9.0 m s–1. The driver sees a stationary car
ahead and needs to stop. The car decelerates at 2.0 m s–2 when the brakes are applied.
Calculate the braking distance.