Question

a bus travelling is brought to rest by applying brakes with a velocity of 54 km/h . it experiences a uniform retardation of -4m/s^2 . Before coming to rest it covers a distance of what?

Answer 1

$\red{Acceleration\:is\:constant\:}$So,we can use equation of motion.

☆Convert velocity in m/sec.

$54 \times \frac{1000}{3600} \\ = 15$
$\boxed{velocity\: is \:15m/s.}$

By using equation of linear motion-
${v}^{2} - {u}^{2} = 2as$
Finally body have to stop so v=0

$0 - {(15)}^{2} = 2( - 4)s \\ 225 = 8s \\ s = \frac{225}{8} m \\ 28.125m$
$<b>$ before coming to rest body will cover 28.125metres.

Answer 2

V= 0

U = 54kmph = 54*5/18 = 15m/s

A = - 4m/s^2

S =?

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

$(0) {}^{2} - (15) {}^{2} = 2( - 4)s$

$- 225 = - 8s$

$28.12m = s$



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