Kinematics Questions and Answers

Answer: A

The straight-line graphs are A, B and D.

• A: a body at rest keeps the same distance, so its distance-time graph is a horizontal straight line.
• B: with uniform acceleration, velocity increases by equal amounts in equal times, so the velocity-time graph is a sloping straight line.
• D: a body at rest has zero velocity throughout, a straight line along the time axis.

C is not straight: under uniform acceleration distance grows as the square of time (s = ut + ½at²), so the distance-time graph is a curve (a parabola). Options containing C are therefore wrong.

Answer: B

At the highest point the velocity is zero and the gravitational force acts.

On the way up gravity slows the ball until, at the top, its velocity is momentarily zero; then it starts to fall. But the force of gravity does not switch off at the top. The ball's weight, mg, acts downward throughout the flight, which is why it immediately begins to come back down with acceleration g.

So 'net force is zero' is the classic misconception: zero velocity does not mean zero force. And the velocity is at its minimum, not its maximum, at the top.

Answer: A

Neena's average speed is 6 km/h.

Average speed is total distance divided by total time, and the time spent stopped counts.

\[ \begin{aligned} \text{first part} &= 12 \times \tfrac{10}{60} = 2 \text{ km} \\ \text{rest} &= 4 - 2 = 2 \text{ km} \\ \text{its time} &= \tfrac{2}{8} \text{ h} = 15 \text{ min} \end{aligned} \]

\[ \begin{aligned} \text{total time} &= 10 + 15 + 15 = 40 \text{ min} \\ \text{average} &= \frac{4}{40/60} = 6 \text{ km/h} \end{aligned} \]

10 km/h is the average of 12 and 8, which ignores both the stop and the time spent at each speed. 8 km/h is only the speed of the second part, and 4 km/h would need a total time of one hour.

Answer: D

The correct statement is that the velocity-time graph is a straight line parallel to the X-axis.

Uniform velocity means the velocity does not change with time, so a graph of velocity against time is a horizontal line at the value of that velocity, parallel to the time (X) axis.

The distance-time graph for uniform velocity is a straight line sloping upward, with slope equal to the velocity. A distance-time line parallel to the X-axis would mean the object is at rest, and a line parallel to the Y-axis, on either graph, would mean a change at a single instant, which is impossible.

Answer: A

An odometer measures the distance travelled by the car.

It counts the rotations of the wheels or transmission and converts them to the total distance covered, shown in kilometres.

The speedometer shows the instantaneous speed. Average speed is total distance divided by time and is not shown by the odometer itself, and acceleration is not measured by either instrument.

Answer: B

The acceleration is 3 m s^-2.

Acceleration is the time derivative of velocity:

\[ \begin{aligned} \vec{a} &= \frac{d}{dt}\left(2.0\hat{i} + 3.0t\hat{j}\right) \\ &= 3.0\hat{j} \end{aligned} \]

The constant \(x\)-component contributes nothing, so the magnitude is 3. Adding the coefficients 2 and 3 gives 5, and 2 is the constant velocity component, not an acceleration.

Answer: D

The pair is 7 units and 2 units.

The resultant of vectors of sizes \(a\) and \(b\) can take any value from \(|a - b|\) (opposite directions) to \(a + b\) (same direction). For 7 and 2 that range is 5 to 9, which excludes 4.

For the other pairs 4 is reachable: 2 and 2 give 0 to 4; 1 and 3 give 2 to 4; 5 and 1 give 4 to 6; 5 and 8 give 3 to 13.

Answer: C

The ratio is 1 : 60.

The minute hand turns once in 60 minutes (3600 s) and the second hand once in 60 s. Angular speed is \(\omega = \frac{2\pi}{T}\), so

\[ \begin{aligned} \frac{\omega_{m}}{\omega_{s}} &= \frac{T_{s}}{T_{m}} \\ &= \frac{60}{3600} = \frac{1}{60} \end{aligned} \]

60 : 1 inverts the ratio, and 1 : 12 compares the minute hand with the hour hand.

Answer: D

\(t_{1}\) is equal to \(t_{2}\).

Without air resistance the motion is symmetric. Going up from speed \(u\) to rest takes \(t_{1} = \frac{u}{g}\), reaching height \(h = \frac{u^{2}}{2g}\). Falling back through the same \(h\) from rest takes \(t_{2} = \sqrt{\frac{2h}{g}} = \frac{u}{g}\).

So the times are equal. The multiples 0.5, 0.25 and 4 would need a different acceleration on the way up and down, which happens only with air resistance (then ascent is shorter than descent).

Answer: B

The object passes the height \(h/2\) at \(\frac{t}{\sqrt{2}}\).

Falling from rest, the distance fallen is \(s = \frac{1}{2}gt^{2}\). At height \(h/2\) above the ground it has fallen \(h/2\):

\[ \begin{aligned} h &= \tfrac{1}{2}gt^{2} \\ \tfrac{h}{2} &= \tfrac{1}{2}gt'^{2} \\ t' &= \frac{t}{\sqrt{2}} \end{aligned} \]

The distance grows with the square of the time, so half the distance takes \(1/\sqrt{2}\) of the time, not half (option C). \(\sqrt{2}t\) and \(2t\) are longer than the whole fall.