# How long will the acceleration of 10 m/2 take place if the bus will run 25m/s and will change to 15 m/s?

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The velocity changes from 25 m/sec to 15 m/sec. This represents a change of 10 m/sec. An acceleration of 10m/sec/sec implies that there will be a change of 10 m/sec for every second the acceleration acts. So change in velocity = acceleration times time. So, in this case, t = 1.0 seconds.

initial velocity=u=25m/s;final velocity=v=15m/s

acceleration=a=10m/s^2

a=(v-u)/t

=>t=-a/(v-u)

=-10/(15-25)

= -10/-10

= 1 sec

Because the velocity is decreased,hence acceleration is negative.

The first answer is incorrect as the final velocity is given as 15m/s instead of 10 m/s..

initial velocity=u=25m/s;final velocity=v=15m/s

acceleration=a=10m/s^2

a=(v-u)/t

=>t=-a/(v-u)

=-10/(15-25)

= -10/-10

= 1 sec

Because the velocity is decreased,hence acceleration is negative.

Given acceleration a = -10 m/s^2, initial velocity u = 25 m/s, and

final velocity v = 15 m/s.

Please note that accelaration is taken as negative because te speed decreases.

Then time taken for change in velocity t = (v - u)/a

= (15 - 25)/(-10) = 1

Answer: The acceleration should take place for 1 s.

The acceleration of 10 m/s^2 should act in opposing direction of the velocity of the bus in order that the velocity (speed) of 25m/s comes down to 15m/s.

We know that acceleration is the rate of change of velocity.

Therefore, acceleration =( Final velocity -initial velocity )/time

Therefore -10m/s^2 = [(15-25)m/s]/t

t= (15-25)/(-10 )secs

=10/10

=1 secs.

Therefore the acceleration of 10 m/s^2 should act for 1 seconds in a direction opposite to the velocity of the bus.