Question

when brakes are applied to a bus the retardation produced is 25 cm per second and the bus takes 20 seconds to stop calculate the initial velocity of bus and the distance travelled by the bus during this time

Answer 1

a= -25 cm sq.
t=20 sec
v=0 cm/sec.
v=u+at
0=u+(-25)×20
u=500 cm/sec.
now use equation third
v sq.=u sq.+ 2as
0=500×500+2×(-25)×S
s=5000 cm
hence initial velocity is 500 cm/sec. and distance travelled by bus is 5000 cm

Answer 2

Given:

when brakes are applied to a bus the negative acceleration produced is 25 cm per second and the bus takes 20 seconds to stop.

To Find:

The initial velocity of the bus and the distance traveled by bus during this time?

Explanation:

• By the first equation of motion.

        $\bf v=u+at$

        Where v is the final velocity, u is the initial velocity,  a is

        the acceleration and t is the time taken.

• In this given problem, the negative acceleration produced by the bus is 25 cm/second and it takes 20 seconds to stop.

        $a=-25cm/s$ $, t=20s$

        put these values in the first equation of motion.

        $0=u-(0.25)\times (20)$

        $\bf u=5m/s$

• By second equation of motion

        $\bf S=ut+\frac{1}{2} at^2$

        Where s is the distance.

        Put the values to find the distance traveled.

        $s=5\times(20)-\frac{1}{2} (0.25)(20)^2$

        $s=100-0.25\times200$

        $s=100-50=50m$

So, the initial velocity of bus is 5m/s and distance travelled during this time is 50m.