Question

A car is moving with velocity 108 km/h. If driver of the car want to
change its velocity from 108 km/h to 90 km/h in two minutes.
Mass of the car is 300 kg. What force applied by the brakes on the
tyres?
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Answer 1

Answer:

108km/h = 108×5/18 m/s = 30m/s

90km/h = 90×5/18 = 25m/s

a = (u - v)/t

a = (30 - 25)/(2 × 60)

a = 1/24 m/s²

F = ma

F = 300 × 1/24

F = 12.5 N

Force applied by the brakes on the tyres is 12.5N.

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Answer 2

Answer:

A car is moving at a velocity of 108 km/h. If the driver of the car wants to change its velocity from 108 km/h to 90 km/h in two minutes.

The mass of the car is 300 kg. The force applied by the brakes on the

tyres will be $12.5\ N$        

Explanation:

Force (F) applied by the breaks on the tyres can be calculated by using Newton's second law of motion Which is stated as,

$F = ma$     ...(1)    

Where,

m -  the mass of the object

a - acceleration

Acceleration can be obtained by calculating the rate of change of velocity with respect to time.

That is,

$a = \frac{v-u}{t}$    ...(2)

where

u -  Initial velocity

v -  Final velocity

t  -  Time

From the question,

$u = 108\ km/h = 108\times \frac{5}{18} = 30 m/s$

$v = 90\ km/h = 90\times \frac{5}{18} = 25 m/s$

$m = 300\ kg$

$t = 2\ min = 2\times 60 = 120\ s$

Substituting these values into equation (2),

$a = \frac{25-30}{120} = -\frac{1}{24} \ m/s^{2}$ ....(3)

The negative sign indicates the decrease in velocity. Negative acceleration is called deceleration or retardation.

Substitute the value of 'a' into equation (1)

$F = 300\times \frac{1}{24} = 12.5\ N$

So the force applied by the break  $= 12.5\ N$