Kinematics Questions and Answers

Answer: B

The quantity is the average velocity.

The change in position, x₂ − x₁, is the displacement. Displacement divided by the time interval over which it happens is, by definition, the average velocity over that interval. It depends only on the end points, not on how the body moved in between.

Instantaneous velocity is the limit of this ratio as Δt tends to zero, the derivative dx/dt, so it needs a vanishing interval, not a finite one. Average and instantaneous acceleration are rates of change of velocity, not of position. 'Average displacement' is not a rate at all, and the ratio has units of m s^-1, not metres.

Answer: E

The car travels 7200 m.

Phase 1, speeding up from rest at \(2\ \text{m s}^{-2}\) for 60 s:

\[ \begin{aligned} v &= at = 2 \times 60 = 120\ \text{m s}^{-1} \\ s_{1} &= \tfrac{1}{2}at^{2} = \tfrac{1}{2} \times 2 \times 60^{2} \\ &= 3600\ \text{m} \end{aligned} \]

Phase 2, slowing from 120 m/s to rest at the same rate in another 60 s, is the mirror image, so it covers another 3600 m:

\[ s = 3600 + 3600 = 7200\ \text{m} \]

The trap is 3600 m, which counts only one phase. A velocity-time graph is a quick check: a triangle with base 120 s and height 120 m/s has area \(\tfrac{1}{2} \times 120 \times 120 = 7200\) m.

Answer: A

The height difference is 80 m.

Falling from rest, the distance fallen in time \(t\) is \(\frac{1}{2}gt^{2}\):

\[ \begin{aligned} h_{3} &= \tfrac{1}{2}(10)(3)^{2} = 45\ \text{m} \\ h_{5} &= \tfrac{1}{2}(10)(5)^{2} = 125\ \text{m} \\ h_{5} - h_{3} &= 80\ \text{m} \end{aligned} \]

Using \(\frac{1}{2}g(5 - 3)^{2} = 20\) m treats the motion as if it restarted from rest, and the other options come from using \(gt\) or wrong times.

Answer: C

\(\tan\theta =\) 4.

Set the range equal to the maximum height:

\[ \begin{aligned} \frac{2u^{2}\sin\theta\cos\theta}{g} &= \frac{u^{2}\sin^{2}\theta}{2g} \\ 4\cos\theta &= \sin\theta \;\Rightarrow\; \tan\theta = 4 \end{aligned} \]

(about \(76^{\circ}\)). \(\tan\theta = 1\) gives the maximum range, at \(45^{\circ}\), not equal range and height, and \(\frac{1}{2}\) or \(\sqrt{2}\) come from dropping a factor of 2.

Answer: B

The swimmer's speed in still water is 8 kmh⁻¹.

Downstream the speeds add and upstream they subtract: \(v + u = 10\) and \(v - u = 6\). Adding:

\[ \begin{aligned} 2v &= 16 \\ v \\ &= 8\ \text{km h}^{-1}, \quad u \\ &= 2\ \text{km h}^{-1} \end{aligned} \]

2 is the river's speed, the usual confusion, and the other values do not satisfy both equations.

The general result: still-water speed is the average of the downstream and upstream speeds, and the river's speed is half their difference.

Answer: C

The ratio of orbital velocities is 2 : 5.

Speed in a circle is \(v = 2\pi rf\). A completes 1 circle while B completes 5, so \(f_{A} : f_{B} = 1 : 5\), and \(r_{A} : r_{B} = 2 : 1\):

\[ \begin{aligned} \frac{v_{A}}{v_{B}} &= \frac{r_{A}f_{A}}{r_{B}f_{B}} \\ &= \frac{2 \times 1}{1 \times 5} \\ &= \frac{2}{5} \end{aligned} \]

5 : 2 inverts it, 1 : 5 ignores the radii, and 1 : 1 or 1 : 2 do not follow.

Answer: D

The distances are 30 m and 80 m.

Use \(s = ut + \frac{1}{2}at^{2}\) with u = 1 m/s and a = 2 m/s\(^{2}\). The second interval is the distance in 10 s minus the distance in 5 s.

\[ \begin{aligned} s_{5} &= 1(5) + \tfrac{1}{2}(2)(25) = 30\ \text{m} \\ s_{10} &= 1(10) + \tfrac{1}{2}(2)(100) = 110\ \text{m} \\ s_{10} - s_{5} &= 80\ \text{m} \end{aligned} \]

30 m and 110 m gives the total distance after 10 s instead of the distance in the second interval. The other pairs come from leaving out the ut term or the factor \(\frac{1}{2}\).

Answer: E

The ball will fall at 250 m.

The horizontal range is \(R = \frac{u^{2}\sin 2\theta}{g}\). At the same angle, R is proportional to \(u^{2}\).

\[ \begin{aligned} R_{2} &= 160 \times \left(\frac{50}{40}\right)^{2} \\ &= 160 \times \frac{25}{16} = 250\ \text{m} \end{aligned} \]

Scaling with u instead of \(u^{2}\) gives 200 m, which is not offered; 180, 280, 300 and 480 m do not follow from either scaling.

Answer: B

The average velocity is 0.5 m s\(^{-1}\).

Average velocity is displacement divided by time, not distance. A and B are ends of two perpendicular radii, so the displacement AB is the hypotenuse of a right triangle with legs 0.5 m.

\[ \begin{aligned} AB &= \sqrt{0.5^{2} + 0.5^{2}} = 0.5\sqrt{2}\ \text{m} \\ v_{avg} &= \frac{0.5\sqrt{2}}{\sqrt{2}} = 0.5\ \text{m s}^{-1} \end{aligned} \]

Using the arc length \(\frac{\pi}{4}\) m instead gives the average speed, about 0.56 m/s, which is not the average velocity. 0.25 takes only the radius over twice the time.

Answer: A

The velocity and displacement are 48 m/s and 64 m.

Integrate the acceleration twice, starting from rest at the origin:

\[ \begin{aligned} v &= \int_{0}^{t} 6t\,dt = 3t^{2} = 3(16) = 48 \\ s &= \int_{0}^{t} 3t^{2}\,dt = t^{3} = 64 \end{aligned} \]

The other pairs use constant-acceleration formulas, which do not apply when the acceleration varies with time.