Question

A car of mass 1000kg is moving with a
velocity of 36km/hr. On the way the driver
finds a kid playing on the road 25m ahead
and presses the brakes. Find the retarding
force of the brakes of the car so that no
accidents take place.

​

Answer 1

Explanation:

Given,

Speed=20m/s,d=50m,retardation=5m/s

2

Let the ppossible reaction time be t

s

So the car moves 20tm And applied breaks with retardation 5m/s

2

So, the total distance under retardation is ≤50m

Thus the distance under the application of retardation =

2×5

20

2

−0

2

=s

′

⇒s

′

=40m

Therefore, 20t

s

+40≤=50⇒20t

s

≤50−40⇒20t

s

=10⇒t

s

≤

2

1

≤0.5s

Answer 2

Given :

• Initial velocity, u = 36 km/h
• Final velocity, v = 0 m/s (as it applies brakes)
• Distance, s = 25 m

To find :

• Acceleration, a &
• Force, F

According to the question,

Note : At first we have to change 36 km/h into m/s.

So,

⇒ 36 km/h = 36 × 1000/3600 = 10 m/s

Now,

By using Newtons third equation of motion,

⇒ v² = u² + 2as

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

⇒ Substituting the values,

⇒ (0)² = (10)² + 2 × a × 25

⇒ 0 = 100 + 50a

⇒ 0 - 100 = 50a

⇒ - 100 = 50a

⇒ - 100 ÷ 50 = a

⇒ - 2 = a

So,the acceleration is - 2 m/s².

Note : Negative signs means retardation.

Now,

⇒ Force = Mass × Acceleration

Or,

⇒ F = ma

⇒ Substituting the values,

⇒ F = 1000 × (-2)

⇒ F = - 2000 N

.°. The force is - 2000 Newton.

Note : Negative signs means that the force is acting on opposite direction.

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