Equations of Motion

Practice Questions:

Question 1: difficult

A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/hr and 40 km/hr respectively. The velocity of the car midway between P and Q is :

1. 33.3 km/hr

2. \(20\sqrt{3} km / hr\)

3. /(25\sqrt{2} km / hr/)

4. 35 km/hr

View Answer

For uniform acceleration, the velocity midway between two points is given by the formula: \(v_{mid} = \sqrt{\frac{v_1^2 + v_2^2}{2}} = \sqrt{\frac{30^2 + 40^2}{2}} = \sqrt{\frac{2500}{2}} = 25\sqrt{2}\text{ km/hr}\).

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Question 2: difficult

A paratrooper jumps from a height \(H\). The parachute can provide a uniform deceleration of \(2\text{ ms}^{-2}\). The height above the ground at which the parachute should be opened so that he touches ground with zero speed is (take \(g = 10\text{ ms}^{-2}\)):

1. \(\frac{H}{6}\)

2. \(\frac{4H}{5}\)

3. \(\frac{5H}{6}\)

4. \(\frac{6H}{7}\)

View Answer

Let \(h\) be free fall and \(y\) be decelerating height, so \(H = h + y\). Speed before parachute opens: \(v^2 = 2gh = 20h\). Deceleration phase: \(0 = v^2 - 2ay = v^2 - 4y\), which gives \(20h = 4y \Rightarrow y = 5h\). Since \(H = 6h\), we find \(y = \frac{5H}{6}\).

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