Question

A bus is moving with an initial velocity 'u' m/s . After applying brakes its retardation is 0.5m/s2 and is stopped after 12s .Find the initial velocity 'u' and distance travelled by the bus after applying brakes ? ​

Answer 1

Answer:

Final velocity   v=0

Using    v=u+at

So,    0=u+(−0.5)×12

⟹ u=6 m/s

Using,   v2−u2=2aS

Or,     0−62=2×(−0.5)×S

⟹ S=36 m

Explanation:

Answer 2

Given :

• Initial Velocity of bus = u
• Acceleration = - 0.5 m/s²
• Final velocity = 0 ( brakes applied )
• Time = 12 sec

To Find :

• Initial Velocity of bus
• Distance travelled by the bus after applying brakes.

Solution :

Part 1:

We have to find initial velocity of the bus

By First equation of motion :

$\bf\:v=u+at$

Put the given values

$\sf\implies\:0=u+(-0.5)(12)$

$\sf\implies\:u=0.5\times12$

$\sf\implies\:u=6ms^{-1}$

Therefore , initial velocity of the bus is 6 m/s

Part -2 :

We have to find distance travelled by the bus after applying brakes.

By third equation of motion

$\bf\:v{}^{2}=u{}^{2}+2aS$

Put the given values

$\sf\implies\:0=6^2+2(-0.5)\times\:S$

$\sf\implies\:S=\dfrac{6^2}{2\times0.5}$

$\sf\implies\:S=36m$

Therefore,Distance travelled by the bus after applying brakes is 36m.

___________

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion.

$\bf\:v=u+at$

$\bf\:S=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2aS$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$