Kinematics Questions and Answers

Answer: B

\(x = 5\).

The scalar product adds the products of matching components:

\[ \begin{aligned} 2x + 3(-3) + 2(4) &= 9 \\ 2x - 1 = 9 &\Rightarrow x = 5 \end{aligned} \]

Forgetting the negative sign on \(-3\hat{j}\) gives \(2x + 17 = 9\), a negative \(x\). The value 9 confuses the given product with \(x\).

Answer: B

The particle reverses its direction of motion at that instant.

Where the v-t graph cuts the time axis, the velocity is zero and changes sign, from positive to negative or the other way. A change in the sign of velocity is a reversal of the direction of motion, as at the top of a ball's flight.

The displacement there is only a turning point: a maximum if the velocity goes from positive to negative, a minimum if it goes the other way, so 'maximum displacement' is not true in general. The velocity is zero, not maximum, and it is changing, not constant. The graph's crossing says nothing about the acceleration being maximum.

Answer: A

B overtakes A after 5 s.

Both start together from the same point, so B overtakes A when they have covered equal distances.

\[ \begin{aligned} 5t &= \tfrac{1}{2} \times 2 \times t^{2} = t^{2} \\ t &= 5\ \text{s} \end{aligned} \]

(The other root, t = 0, is the start.) At 5 s both have gone 25 m, and B's speed is then 10 m/s, twice A's. 10 s is the time when B has twice A's distance, not when it draws level.

Answer: B

The area under a velocity-time graph equals the displacement.

Over a short time dt the displacement is v dt, the area of a thin strip under the graph. Adding all the strips, the total area between the graph and the time axis is the total displacement, with areas below the axis counted as negative.

Acceleration is the slope of the velocity-time graph, not its area. Velocity and speed are the values read off the graph, and angular velocity belongs to rotation. (Distance travelled is the area if parts below the axis are counted as positive.)

Answer: B

The angle is 45°.

The angle \(\theta\) with the x-axis satisfies \(\tan\theta = \frac{A_{y}}{A_{x}}\). With equal components,

\[ \tan\theta = 1 \Rightarrow \theta = 45^{\circ} \]

30° and 60° need components in the ratio \(1 : \sqrt{3}\). 90° means no x-component, and 0° means no y-component.

Answer: B

The velocity is zero at 2 s.

Velocity is the time derivative of position:

\[ \begin{aligned} v &= \frac{d}{dt}(t^{3} - 3t^{2} + 2) = 3t^{2} - 6t \\ 3t(t - 2) &= 0 \Rightarrow t = 0 \text{ or } t = 2\ \text{s} \end{aligned} \]

\(t = 0\) is the starting instant, so the time during the motion at which the velocity becomes zero is 2 s. The other times do not make \(3t^{2} - 6t\) zero; at 1 s, for example, the velocity is -3.

Answer: C

The displacement is \(10\sqrt{5}\) km.

In the first hour the body covers \(20 \times 1 = 20\) km. It then turns through a right angle and moves for half an hour at the same speed, covering \(20 \times 0.5 = 10\) km.

\[ \begin{aligned} d &= \sqrt{20^{2} + 10^{2}} = \sqrt{500} \\ &= 10\sqrt{5} \text{ km} \end{aligned} \]

30 km would be the distance travelled, not the displacement, and 20 km or 10 km are the two legs on their own.

Answer: B

The linear speed increases by 2.4 m s^-1.

Linear speed is \(v = r\omega\), so with \(r\) fixed, \(\Delta v = r\Delta\omega\).

\[ \Delta v = 1.2 \times (4 - 2) = 2.4 \text{ m s}^{-1} \]

4.8 m s^-1 is the final speed, \(1.2 \times 4\), not the increase, and 2.4 is also the initial speed, which is why the speed exactly doubles.

Answer: C

The angle is 60°.

Differentiate the position vector to get the velocity:

\[ \begin{aligned} \vec{v} &= \frac{d\vec{r}}{dt} = 2\hat{i} + 2\sqrt{3}t\hat{j} \\ \vec{v}(1) &= 2\hat{i} + 2\sqrt{3}\hat{j} \end{aligned} \]

The velocity has no z-component, so it lies in the x-y plane, and its angle with the x-axis satisfies \(\tan\theta = \frac{2\sqrt{3}}{2} = \sqrt{3}\), giving \(\theta = 60^\circ\).

30° is the angle with the y-axis. Using the position vector instead of the velocity would bring in the constant 5 in the z-direction, which does not affect the velocity.

Answer: D

The ball lands 30 m from the foot of the tower.

The vertical motion starts from rest, so the time of fall depends only on the height.

\[ \begin{aligned} 45 &= \tfrac{1}{2}(10)t^{2} \Rightarrow t = 3\ \text{s} \\ x &= 10 \times 3 = 30\ \text{m} \end{aligned} \]

The horizontal speed stays 10 m/s throughout. 45 m is the height, not the range, and 20 m or 40 m come from a wrong time of flight.