Question

A car is travelling at a speed of 90 km / hr. Brakes are applied so as to produce a uniform acceleration of -0.5 m / s². Find how far the car will go before it is brought to rest

Answer 1

GiveN :

Initial velocity (u) = 90 km/h = 25 m/s

Final velocity (v) = 0 m/s

Acceleration (a) = - 0.5 m/s²

To FinD :

Distance travelled by car after applying breaks.

SolutioN :

Use 3rd equation of Kinematics :

⇒v² - u² = 2as

⇒0² - 25² = 2 * -0.5 * s

⇒0 - 625 = -1 * s

⇒s = 625 m

Answer 2

Answer:

Explanation:

First Convert 90km/hr into m/s

=>90km/1hr=90000m/3600 =25m/s

The car's initial velocity is 25m/s.

The car's final velocity=0 [As it stops after applying brakes]

From the formula v=u+at

=>t=  v-u  

          -a(Retardation)

=> t= 0-25

         -0.5

=> t= -25÷-5/10

=> t= -25×-10/5

=> t= 50s

From the formula S=ut+1/2*at²

=> S= 90*50+1/2*0.5*2500(50²)

=> S= 4500+625=5125

Distance =5125m or 5.125km

pls mark as brainliest