Kinematics Questions and Answers

Answer: A

The angular velocity is \(\sqrt{10}\) rad/s.

The centripetal acceleration is \(\omega^{2}r\). Setting it equal to g with \(r = 0.98\) m:

\[ \begin{aligned} \omega^{2} &= \frac{g}{r} \\ &= \frac{9.8}{0.98} \\ &= 10 \\ &\Rightarrow \omega \\ &= \sqrt{10} \end{aligned} \]

10 is \(\omega^{2}\) without the square root, \(\frac{1}{\sqrt{10}}\) inverts the ratio, and 100 or 0.1 come from using the radius in centimetres.

Note the units: the radius must be in metres (0.98 m) to match g in m s⁻².

Answer: D

The displacement varies as \(s \propto t^{2}\).

For uniform acceleration from rest (u = 0), the equation of motion \(s = ut + \frac{1}{2}at^{2}\) reduces to

\[ s = \frac{1}{2}at^{2} \]

With a constant, s is proportional to \(t^{2}\): doubling the time quadruples the distance. \(s \propto t\) describes uniform velocity, not uniform acceleration. \(s \propto t^{1/2}\) inverts the relation (it is t that varies as \(\sqrt{s}\)), and the negative powers would make the displacement shrink with time.

Answer: C

The quantity that stays constant is the speed.

Covering equal distances in equal times means uniform circular motion: the magnitude of the velocity is constant. But its direction keeps turning along the tangent to the circle.

So the velocity, a vector, changes all the time, and with it the linear momentum. The acceleration has constant magnitude v²/r but always points to the centre, so its direction changes too. The displacement from the starting point changes as the particle moves round. Only the speed, a scalar, is constant.

Answer: C

The angle is \(\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\).

The angle a vector makes with the \(y\)-axis is given by its direction cosine along \(\hat{j}\): the \(y\)-component divided by the magnitude.

\[ \begin{aligned} |\vec{A}| &= \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3} \\ \cos\beta &= \frac{A_{y}}{|\vec{A}|} = \frac{1}{\sqrt{3}} \end{aligned} \]

So \(\beta = \cos^{-1}\frac{1}{\sqrt{3}} \approx 54.7^{\circ}\). \(\sin^{-1}\frac{1}{\sqrt{3}}\) is the complementary angle, made with the \(xz\)-plane. \(\frac{2}{\sqrt{3}}\) is greater than 1, so neither \(\cos^{-1}\) nor \(\sin^{-1}\) of it exists. \(\frac{\pi}{2}\) would need \(A_{y} = 0\).

Answer: D

The graph is an oblique straight line with negative slope.

The slope of a velocity-time graph is the acceleration. Constant retardation means a constant negative acceleration, so \(v = u - at\) is a straight line starting at \(u\) (nonzero) and falling steadily.

A positive slope would be uniform acceleration. A line parallel to the time axis is constant velocity (zero acceleration). A line parallel to the velocity axis would mean an instant change of speed, which is not physical. A curve would mean the acceleration itself is changing.

Answer: E

\(\frac{T_{1}}{T_{2}} = \tan\theta\).

The time of flight of a projectile launched at speed \(u\) and angle \(\alpha\) is \(T = \frac{2u\sin\alpha}{g}\). For the angles \(\theta\) and \(90^{\circ} - \theta\):

\[ \begin{aligned} \frac{T_{1}}{T_{2}} &= \frac{\sin\theta}{\sin(90^{\circ} - \theta)} \\ &= \frac{\sin\theta}{\cos\theta} \\ &= \tan\theta \end{aligned} \]

These complementary angles give the same range, but not the same time of flight, because the time depends only on the vertical component of velocity. \(\cot\theta\) is the ratio \(T_{2}/T_{1}\), and the single sine or cosine options ignore one of the two times.

Answer: A

The acceleration is 6 m s^-2.

Differentiate the position twice:

\[ \begin{aligned} \vec{v} &= \frac{d\vec{r}}{dt} = 8\hat{i} + 6t\hat{j} \\ \vec{a} &= \frac{d\vec{v}}{dt} = 6\hat{j} \end{aligned} \]

The magnitude is 6. 3 is the coefficient of \(t^{2}\) without differentiating, and 8 is the constant velocity component.

The \(x\)-component \(8t\) gives a constant velocity and the \(z\)-component 3 is fixed, so only the \(y\)-motion is accelerated.

Answer: E

The average velocity is 24 km h^-1.

For equal distances \(d\) at speeds \(v_{1}\) and \(v_{2}\), the average is total distance over total time:

\[ \begin{aligned} \bar{v} &= \frac{2d}{\frac{d}{v_{1}} + \frac{d}{v_{2}}} = \frac{2v_{1}v_{2}}{v_{1} + v_{2}} \\ &= \frac{2 \times 20 \times 30}{50} = 24 \end{aligned} \]

The trap is the arithmetic mean, 25 km h^-1. The car spends longer at the slower speed, so the true average, the harmonic mean, is lower than 25.

Answer: D

The vector is \(\frac{1}{\sqrt{5}}(3\hat{j} - 4\hat{k})\).

The required vector is the unit vector along \(\vec{A}\) times the magnitude of \(\hat{i} - 2\hat{j}\).

\[ \begin{aligned} |\vec{A}| &= \sqrt{9 + 16} = 5, \quad \hat{A} = \tfrac{1}{5}(3\hat{j} - 4\hat{k}) \\ |\hat{i} - 2\hat{j}| &= \sqrt{5} \\ \text{vector} &= \tfrac{\sqrt{5}}{5}(3\hat{j} - 4\hat{k}) = \tfrac{1}{\sqrt{5}}(3\hat{j} - 4\hat{k}) \end{aligned} \]

The option \(\sqrt{5}(3\hat{j} - 4\hat{k})\) forgets to make \(\vec{A}\) a unit vector first, giving magnitude \(5\sqrt{5}\). The options containing \(\hat{i}\) are not parallel to \(\vec{A}\), which has no \(\hat{i}\) component.

Answer: C

The ratio is 9 : 8.

The vertical component of velocity is \(u\sin 30^{\circ} = 15\) m s^-1.

\[ \begin{aligned} H &= \frac{(u\sin\theta)^{2}}{2g} = \frac{225}{20} = 11.25\ \text{m} \\ h_{1} &= 15(1) - \tfrac{1}{2}(10)(1)^{2} = 10\ \text{m} \\ H : h_{1} &= 11.25 : 10 = 9 : 8 \end{aligned} \]

The body reaches its maximum height at \(t = 1.5\) s, so after 1 s it is still rising. Using the full 30 m s^-1 as the vertical speed gives very different heights that match none of the ratios.