Motion in One Dimension — Question 11

As we know,

S = ut + $\dfrac{1}{2}$ at²

Given,

u = 0 m s ^-1

S = 270 m

t = 3 s

Substituting the values in the formula above we get,

$270 = (0 \times 3) + (\dfrac{1}{2} \times \text a \times 3^2) \\[0.5em] 270 = \dfrac{9}{2} \times \text a \\[0.5em] 270 \times 2 = 9 \times \text a \\[0.5em] \Rightarrow \text a = \dfrac{540}{9} \\[0.5em] \Rightarrow \text a = 60\ \text{ms}^{-2}$

Hence, a = 60 m s^-2

To find out velocity after t = 10 s.

v = u + at

Substituting the values in the formula above we get,

$\text v = 0 + (60 \times 10) \\[0.5em] \Rightarrow \text v = 600\ \text{ms}^{-1} \\[0.5em]$

Hence, velocity after 10 s = 600 m s^-1