Question

pls answer this as soon as possible.....​

Answer 1

Answer:

Initial velocity of car, u = 108km/h

108 * 1000/3600

108*10/36

3*10

30m/s

Now,

From position-velocity relation,

v^2 - u^2 = 2as

(30)^2 - 0 = 2*a*150

900/2 = 150a

450 = 150a

a = 450/150

a = 3m/s^2

OPTION C IS THE RIGHT ANSWER

HOPE IT HELPS

PLZ DO MARK ME AS THE BRAINLIEST I REALLY NEED IT

Answer 2

Answer:

3.00 m/s²

Explanation:

Initial velocity of the car (u) = 108km/hr = 30m/s

Distance travelled before coming to rest (s) = 150 m.

Final velocity of the car (v) = 0 (as it comes to rest)

Using the 3rd equation of motion,

v^2 - u^2 = 2as

a = (v² - u²)/2s

=> (0 - 30²)/2 * 150

=>  -900/300

= -3 m/s²

Therefore,

Retardation of the car = 3.00 m s⁻²

Hope it helps!