Motion in a Straight Line - Kinematic Equations for Uniformly Accelerated Motion

A car moving at $20 \text{ m/s}$ is brought to rest by applying brakes over a distance of $40 \text{ m}$. Calculate the uniform acceleration (retardation) of the car.

Given: initial velocity $u = 20 \text{ m/s}$, final velocity $v = 0$ (rest), and displacement $s = 40 \text{ m}$. Using the third equation of motion: $v^2 = u^2 + 2as$. Substituting the values: $0^2 = (20)^2 + 2a(40) \implies 0 = 400 + 80a \implies 80a = -400 \implies a = -5 \text{ m/s}^2$.

The negative sign indicates that the acceleration is in the opposite direction to the velocity, which is termed as retardation or deceleration.

A ball is dropped from a tower of height $122.5 \text{ m}$. How long does it take to reach the ground? (Take $g = 9.8 \text{ m/s}^2$)

Given: $u = 0$ (dropped from rest), $s = 122.5 \text{ m}$, $a = g = 9.8 \text{ m/s}^2$. Using the second equation: $s = ut + \frac{1}{2}at^2$. Substituting the values: $122.5 = 0(t) + \frac{1}{2}(9.8)t^2 \implies 122.5 = 4.9t^2 \implies t^2 = \frac{122.5}{4.9} = 25 \implies t = 5 \text{ s}$.

Since the ball is dropped, the initial velocity is zero. We use the positive value for $g$ assuming the downward direction as positive for the displacement.