Question

a car moving along straight highway with speed 126 km/hr is brought to a stop within distance 200m what is the acceleration & how long does it take for car to stop

Answer 1

$\textbf{ Hello Mate! }$

Here, v ( final velocity ) = 0

u ( Initial Velocity ) = 126 km/h

Converting km/h into m/s = 126 × 5 / 18 m/s
= 35 m/s

s ( distance ) = 200 m

v^2 - u^2 = 2as

0^2 - 35^2 = 2a(200)

- 1225 = 400a

a = - 3.0625 or - 3.1 m/s^2

v = u + at

0 = 35 + (-3.1)(t)

- 35 / - 3.1 = t

11.3 second = time

$\boxed{ Acceleration = - 3.0625 m/s^2 and \: Time = 11.3 s }$

Have great future ahead!

$\textbf{ Happy Independence Day }$

जय हिन्द!

Answer 2

Motion

Here
initial speed, u = 126 km/hr
= 126×5/18 m/s
= 35m/s
final velocity, v = 0 m/s [as it is stopped]
distance travelled, s = 200

let the acceleration be a m/s² and the time taken be t s

Now we know that
v²- u²= 2as
=> 0 - 1225 = 2×200a
=> -1225 = 400a
=> a = -1225/400
=> a = -3.0625 m/s²or -49/16

And

v = u + at
=> 0 = 35 -49/16t
=> t = 35×16/49
=> t = 11.43 s

That's it
Hope it helped
=_=