Motion in One Dimension — Question 2

According to equation of motion —

S = ut + $\dfrac{1}{2}$ at²

Where,

distance covered = S

initial velocity u = 10 m s^-1

acceleration a = 5 m s^-2

time t = 5 s

Substituting the values in the formula, we get,

$\text S = (10 \times 5) + (\dfrac{1}{2} \times 5 \times 5^2) \\[0.5em] \text S = (50) + (\dfrac{1}{2} \times 5 \times 25) \\[0.5em] \text S = (50) + ( 2.5 \times 25) \\[0.5em] \text S = 50 + 62.5 \\[0.5em] \Rightarrow \text S = 112.5\ \text m \\[0.5em]$

Hence, distance covered = 112.5 m.