Question

A bus at adistance of 50 m from child suddenly bus accelerates at 1m/s2 . With what minimum velocity child should be run ti catch bus

Answer 1

Answer:

10 m / sec.

Explanation:

Given :

Distance ( s ) = 50 m

Acceleration ( a ) = 1 m / sec²

Initially velocity u = 0 m / sec

Applying third equation :

v²  = u²  + 2 a s

v²  = 2 × 1 × 50  [ u= 0 ]

v²  = 100

v = √ 100

v = 10 m / sec.

Thus the minimum velocity of child should be 10 m / sec

Answer 2

$\bold{\underline{\underline{Answer:}}}$

•°• Minimum velocity required by child to reach the bus is 10 m/s

$\bold{\underline{\underline{Step\:-\:by\:-\:step\:explanation:}}}$

Given :-

• Distance (s) = 50 m
• Acceleration (a) = 1 m/s

To find :-

• Minimum velocity required by child to reach the bus (v)

Solution :-

Initial Velocity (u) = 0 m/s

[Since the child is at rest]

We have, Distance (s), Acceleration (a) and initial velocity (u), we can thereby use the third equation of motion.

$\implies$ $\bold{v^2\:=\:u^2\:+\:2as}$

$\implies$ $\bold{v^2\:=0^2\:+\:2\times\:1\:\times\:50}$

$\implies$ $\bold{v^2\:=0\:+\:2\times\:50}$

$\implies$ $\bold{v^2\:=0\:+100}$

$\implies$ $\bold{v^2\:=100}$

$\implies$ $\bold{v\:=\:\sqrt{100}}$

v = 10