24. A car is travelling along an expressway at 90 km/h. The driver spots a stalled car and some traffic congestion on the road ahead, and so applies the brakes. The braking action causes a frictional force of 8400 N to act on the 1050 kg car. (a) What is the acceleration of the car when the brakes are applied? (b) How long does it take for the car to stop? (c) How far does the car travel while it is braking?

A. The acceleration of the car when the brakes were applied is –8 m/s².

B. The time taken for the car to stop is 3.13 s.

C. The distance travelled by the car while the brakes were applied is 39.06 m.

A. Determination of the acceleration of car.

Force (F) = –8400 N (opposite direction)

Mass (m) = 1050 kg

Force = mass × acceleration

–8400 = 1050 × a

Divide both side by 1050

a = –8400 / 1050

NOTE: The negative sign indicates that the car is coming to rest.

B. Determination of the time taken for the car to stop.

Initial velocity (u) = 90 Km/h = (90 × 1000)/3600 = 25 m/s

Final velocity (v) = 0 m/s

Acceleration (a) = –8 m/s²

v = u + at

0 = 25 + (–8 × t)

0 = 25 – 8t

Collect like terms

0 – 25 = –8t

–25 = –8t

Divide both side by –8

t = –25 / –8

Therefore, the time taken for the car to stop is 3.13 s

C. Determination of the distance travelled by the car.

Initial velocity (u) = 25 m/s

Final velocity (v) = 0 m/s

Acceleration (a) = –8 m/s²

v² = u² + 2as

0² = 25² + (2 × –8 × s)

0 = 625 – 16s

Collect like terms

0 – 625 = –16

–625 = –165

Divide both side by –16

s = –625 / –16

Therefore, the distance travelled by the car is 39.06 m

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See explanation

Explanation:

We convert the speed of the car to ms-1

90 × 1000/3600 = 25 ms-1

a) from;

F= ma

a= F/m

a= 8400/1050 = 8 ms-2

b) from

v= u - at

v= 0

u = 25 ms-1

a= 8 ms-2

t= ?

t = u/a = 25/8 = 3.1 s

From;

v^2 = u^2 - 2as

v= 0 ms-1

u^2 = 2as

s = u^2/2a

s= (25)^2/2 × 8

s= 625/16

s = 39 m

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