Question

A bus is moving with velocity of 60 km/h when breaks are applied it comes to rest after 2 second .find the distance travelled by it before coming to rest

Answer 1

Velocity = 60 km/h or 50/3 m/s

Time = 2 sec

According to Newton's third law

v^2 = u^2 + 2as

Final velocity will be zero as it come to rest after 2 sec and acceleration will be negative because of retardation

(-u^2) = (-2as)

s = u^2 / 2a

s = u×t / 2 ( acceleration = v / t )

s = 50/3 or 16.66 m

Answer 2

Explanation:

$given$

initial velocity,u = 60kmph

$= \frac{150}{9} {ms}^{ - 1}$

from the equation of motion

$v = u + at$

$0 = \frac{150}{9} + a(2)$

$\frac{ - 150}{9} = 2a$

$a = \frac{ - 25}{3}$

From the equation of motion

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

$0 - ({ \frac{150}{9} })^{2} = 2( \frac{ - 25}{3} )(s)$

$\frac{ - 22500}{81} = \frac{ - 50s}{3}$

$- 150s = - 2500$

$s = \frac{ - 2500}{ - 150}$

$s = 16.66m$

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