Question

A car moving at 72km/h is stopped in 5 seconds.If it retards uniformly find the distance travelled by the car before it stops.Also find its retardation.


plzz answer correctly



it's much important​

Answer 1

Answer:

hi pranay

Step-by-step explanation:

given, v= 72km/h

t=5s

a=14.4

using kinemetics equation

$v {}^{2} = u {}^{2} + 2as$

(72)×(72)= 2(14.4)s

s=37324.8

Answer 2

Given

• Initial velocity (u) = 72km/h
• Final velocity (v) = 0
• Time taken (t) = 5s

Find out

• Distance travelled
• Retardation

Solution

➟ Initial velocity = 72km/h

➟ u = 72 × 5/18

➟ u = 20m/s

★According first equation of motion★

➟ v = u + at

➟ 0 = 20 + a × 5

➟ 5a = - 20

➟ a = - 20/5

➟ a = - 4m/s²

Note : Minus shows car retarded uniformly.

★According third equation of motion★

➟ v² = u² + 2as

➟ (0)² = (20)² + 2 × (-4) × s

➟ 0 = 400 - 8s

➟ 8s = 400

➟ s = 400/8

➟ s = 50m

Hence, distance travelled by car i.e 50m

Additional Information

• s = ut + ½ at² {2nd equation of motion}

• Acceleration → The rate of change in velocity

$\rule{200}3$