Exemplary Solutions
Problem 1: A car starting from rest accelerates uniformly at $2\text{ m/s}^2$ for $10\text{ seconds}$. It then travels at constant speed for $20\text{ seconds}$ and is finally brought to rest in $5\text{ seconds}$ by applying brakes. Find: (i) Maximum velocity reached, (ii) Total distance traveled.
Solution:
Stage 1 (Acceleration): $u = 0, a = 2\text{ m/s}^2, t_1 = 10\text{ s}$.
$$v_{\max} = u + at_1 = 0 + 2(10) = \mathbf{20\text{ m/s}}$$
$$s_1 = ut_1 + \frac{1}{2}at_1^2 = 0 + \frac{1}{2}(2)(10^2) = 100\text{ meters}$$
Stage 2 (Constant Velocity): $v = 20\text{ m/s}, t_2 = 20\text{ s}, a = 0$.
$$s_2 = v \times t_2 = 20 \times 20 = 400\text{ meters}$$
Stage 3 (Retardation): $u = 20\text{ m/s}, v = 0, t_3 = 5\text{ s}$.
$$s_3 = \text{Average velocity} \times t_3 = \frac{20 + 0}{2} \times 5 = 10 \times 5 = 50\text{ meters}$$
Total Distance Traveled:
$$s = s_1 + s_2 + s_3 = 100 + 400 + 50 = \mathbf{550\text{ meters}}$$
Problem 2: A ball is thrown vertically upwards with a velocity of $49\text{ m/s}$. Calculate: (i) The maximum height reached, (ii) The total time taken to return to the ground (take $g = 9.8\text{ m/s}^2$).
Solution:
Initial velocity $u = 49\text{ m/s}$. At maximum height, final velocity $v = 0$. Downward acceleration $a = -g = -9.8\text{ m/s}^2$.
(i) Maximum Height ($h$):
$$v^2 = u^2 - 2gh \implies 0 = 49^2 - 2(9.8)h$$
$$19.6h = 2401 \implies h = \frac{2401}{19.6} = \mathbf{122.5\text{ meters}}$$
(ii) Time of Ascent ($t$):
$$v = u - gt \implies 0 = 49 - 9.8t \implies t = \frac{49}{9.8} = 5\text{ seconds}$$
Total time of flight $= 2 \times t_{\text{ascent}} = 2 \times 5 = \mathbf{10\text{ seconds}}$.
