$A$ cyclist driving at $5\, m s^{-1}$ picks up 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.

(N/A) Given: Initial velocity $u = 5\, m s^{-1}$,Final velocity $v = 10\, m s^{-1}$,Distance $S = 50\, m$.
$(i)$ To calculate acceleration $(a)$,we use the third equation of motion: $v^{2} - u^{2} = 2aS$.
Substituting the values: $(10)^{2} - (5)^{2} = 2 \times a \times 50$.
$100 - 25 = 100a$.
$75 = 100a$.
$a = \frac{75}{100} = 0.75\, m s^{-2}$.
$(ii)$ To calculate time $(t)$,we use the first equation of motion: $v = u + at$.
Substituting the values: $10 = 5 + 0.75 \times t$.
$10 - 5 = 0.75t$.
$5 = 0.75t$.
$t = \frac{5}{0.75} = 6.67\, s$.