Question

A truck is moving with the speed of 72 km/h. Suddenly, driver
sees an obstacle and he applies the brake to stop the truck in
5 sec. Find the retardation produced by brakes and the
distance travelled by the truck before it stops.​

Answer 1

a = -4m/s^-2

s = 50m

explanation

Take u = 20m/s

v = 0 ( bcoz the car stops as break is applied)

t = 5 sec

Comment for any doubt

Thanks

Answer 2

Answer :

• Retardation of truck is 4 m/s²
• distance travelled by truck after applying brakes and before coming to rest is 50 m.

Explanation :

• initial velocity of Truck, u = 72 km/h = 72×(5/18)  m/s = 20 m/s
• final velocity of truck, v = 0    (since it comes to rest)
• time taken by truck to stop after applying brakes, t = 5 sec

we need to find,

• Retardation produced by breaks = ?
• distance travelled by truck after applying breaks, s = ?

Retardation is simply called negative acceleration.

Let, acceleration of truck, a = ?

so,

Using first equation of motion

→ v = u + a t

→ 0 = 20 + a (5)

→ -20 = 5 a

→ a = - 4 m/s²

So, we get acceleration of truck is -4 m/s²

therefore,

• Retardation of truck is 4 m/s².

Now,

using second equation of motion

→ s = u t + 1/2 a t²

→ s = ( 20 ) ( 5 ) + 1/2 ( -4 ) ( 5 )²

→ s = 100 + ( - 50 )

→ s = 50 m

therefore,

• Distance travelled by truck after applying brakes and before coming to rest is 50 m.