Forces and Motion - Equations of Motion

A racing car starts from rest and accelerates uniformly at $5 \text{ m/s}^2$ for $8 \text{ seconds}$. Calculate the final velocity and the total distance traveled.

Given: $u = 0 \text{ m/s}$, $a = 5 \text{ m/s}^2$, $t = 8 \text{ s}$.

To find final velocity $v$: $v = u + at$ $v = 0 + (5 \times 8)$ $v = 40 \text{ m/s}$

To find displacement $s$: $s = ut + \frac{1}{2}at^2$ $s = (0 \times 8) + \frac{1}{2}(5)(8^2)$ $s = 0 + \frac{1}{2}(5)(64)$ $s = 160 \text{ m}$

Since the car starts from rest, $u = 0$. We use the first equation of motion to find the velocity reached after 8 seconds and the second equation to find the distance covered during that acceleration period.

A ball is thrown vertically upwards with an initial speed of $20 \text{ m/s}$. Calculate the maximum height reached by the ball. (Assume $g = -9.8 \text{ m/s}^2$)

Given: $u = 20 \text{ m/s}$, $a = -9.8 \text{ m/s}^2$. At maximum height, final velocity $v = 0 \text{ m/s}$.

Using the formula: $v^2 = u^2 + 2as$ $0^2 = (20)^2 + 2(-9.8)s$ $0 = 400 - 19.6s$ $19.6s = 400$ $s = \frac{400}{19.6}$ $s \approx 20.41 \text{ m}$

At the highest point of a vertical trajectory, the instantaneous velocity is zero. We use the sign convention where upwards is positive, making the acceleration due to gravity negative.