Question

A bus comes to rest by applying brakes in 10 seconds. Calculate the initial velocity of the bus and distance travelled during the retardation of 4 meter per second square.

Answer 1

Answer:

t=10

v=0

u=?

d=?

a=(-4)

Explanation:

Since a= v-u/t

        -4=0-u/10

      -40=-u

Therefore, initial velocity is 40 m/s

and,

s=ut+1/2at^2

s=40x10+1/2x(-4)10^2

s=200x(-400)

s=8m

Answer 2

Answer:

• The Initial velocity (u) is 40 m/s
• The distance (s) travelled is 200 m

Given:

1. Acceleration produced (a) =  - 4 m/s²
2. Time period (t) = 10 seconds.

Explanation:

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Before solving, we should now a basic point.

As question says that the bus come to rest, This implies that the final velocity (v) of the bus is zero. And, the acceleration will be taken as negative because bus is retarding.

Now,

Applying first kinematic equation,

⇒ v = u + a t

Here,

• v Denotes final velocity.
• u Denotes Initial velocity.
• a Denotes acceleration.
• t Denotes time taken.

Substituting the values,

⇒ 0  = u + ( - 4 ) × 10

⇒ 0 = u + ( - 40 )

⇒ 40 = u

⇒ u = 40

⇒ u = 40 m/s

∴ The Initial velocity (u) is 40 m/s.

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Applying second kinematic equation

⇒ s = u t + 1 / 2 a t²

Here,

• s Denotes distance travelled.
• u Denotes initial velocity.
• t Denotes time taken.
• a Denotes acceleration.

⇒ s = (40 × 10) + 1 / 2 × (-4) × (10)²

⇒ s = 400 + (-2) × 100

⇒ s = 400 - 200

⇒ s = 200

⇒ s = 200 m

∴ The distance (s) travelled is 200 m.

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