Question

A car travels at a uniform velocity of 60ms-1 for 10 s and is brought to rest in 5s the retardation of the car and the distance traveled in 15s is Option 1) 10ms-2,1750m 2) 12ms-2,350m 3)15ms-2,650m 4)12ms-2, 750m

Answer 1

Answer:

U= 60 m/s

t = 10 sec

a =0

s= ut +(1/2)at²

= 60×10  +0

= 600 m

now, u = 60 m/s

v =0

t =5 sec

as, v= u+at

⇒0 = 60 + a×5

⇒5a = -60

⇒a = - 12 m/s²

also, s = ut +(1/2)at²          

= 60×5 +(1/2)×(-12)×(5)²          

= 150 m

Total distance = 600+150

                        = 750 m

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Answer 2

Answer:

Explanation:

Distance = speed *time

distance during first 10 seconds = 60 * 10= 600m

acceleration = (change in speed) / time

acceleration during next 5 second = (0-60)/5 = -12m/s^2

by third kinematic equation,

2as = v^2 - u^2

2 (-12) s = (0)- 60^2

24 s = 3600

s = 150m

total distance = 750m