Question

A truck is moving at a speed of 50 Ms-1 .if brakes are applied it regards at a rate if 20 m/min . Find the stopping distance

Answer 1

AnswEr :

Given that a truck is moving with speed of 50 m/s. If brakes are applied it retards at a rate of 20 m/min

We have to find stopping distance of truck.

From the given data we have :-

➡ Initial speed (u) = 50 m/s

➡ Final speed (v) = 0 m/s [As it stops]

➡ Retardation = 20 m/min = 20 m/60 s = 0.33 m/s²

∴ Acceleration = -0.33 m/s²

First let's find time taken to stop :-

We will use 1st equation of motion :-

⇒ v = u + at

Substituting values,

⇒ 0 = 50 + (-0.33)t

⇒ 0 = 50 - 0.33t

⇒ 0.33t = 50

⇒ t = 50/0.33

⇒ t = 151.52

Now we will use 2nd equation of motion :-

⇒ s = ut + ½ at²

Substituting values,

⇒ s = 50 × 151.52 + ½ (-0.33) × (151.52)²

⇒ s = 7576 + ½ × (-0.33) × 22958.31

⇒ s = 7576 - 3788.12

⇒ s = 3787.88 m

∴ Stopping distance of truck = 3787.88 m

Answer 2

Given :-

• Initial velocity =50m/s

• Final velocity =0m/s

• Acceleration of truck = 0.333m/s²(retardation)

To Find:-

• Stopping distance of truck.

Solution :-

• By using 1st equation of motion

➦ v=u+at

➭ 0=50+(-0.333)×t

➭ 0=50+(-0.333)×t

➭ 50= (-0.333)×t

➭ t =50/0.333

➭t = 150.15 s

and Now,

We have to calculate the stopping distance

By applying 2nd equation of motion

☛ s =ut+1/2at²

☛s = 50×150.15+1/2×(- 0.333)×150.15× 150.15

☛s= 7507.5-3753.74

☛s= 3753.76

∴ The stopping distance is 3753.76m