A cyclist driving at $5\, m s ^{-1}$ picks a velocity of $10\, m s ^{-1}$ over a distance of $50\, m$. Calculate $(i)$ acceleration $(ii)$ time in which the cyclist picks up the above velocity.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Given $u=5 m s ^{-1}, v=10 m s ^{-1}, S =50 m , a=?$

$t=?$

Applying $v^{2}-u^{2}=2 a S$

$(10)^{2}-(5)^{2}=2 \times a \times 50$

$75=100 a$

$a=0.75 m s ^{-2}$

$v=u+a t$

$10=5+0.75 \times t$

Therefore, $t=\frac{5}{0.75}=6.67 s$

Similar Questions

If the displacement$-$time graph for a particle is parallel to time axis, what is the velocity of the particle ?

What is meant by uniform motion ? Can you think of an example of a body in uniforim motion ?

Area under speed$-$time graph is equal to the

Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between $4^{th}$ and $5^{th}$ seconds.

If the acceleration of a particle is constant in magnitude but not in direction, what type of path is followed by the particle ?