130 practice questions covering Describing Motion Around Us, organised into 1 topics.
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A stone is released from rest at the top of a tower of known height and falls freely with the uniform acceleration $g$. The time it takes to reach the ground is wanted. Which equation gives that time in one step?
A$s = ut + \frac{1}{2}at^{2}$
B$v^{2} = u^{2} + 2as$
C$a = \frac{v - u}{s}$
D$v = u + at$
Show answer and explanation
Correct answer: A - $s = ut + \frac{1}{2}at^{2}$
The known quantities are $u = 0$, $a = g$ and the height $s$, and the wanted quantity is $t$, so the equation containing all four is $s = ut + \tfrac{1}{2}at^{2}$. The equation $v = u + at$ also contains $t$ but needs the landing velocity, which is not given. The equation $v^{2} = u^{2} + 2as$ has no $t$ in it at all, and dividing a change in velocity by a distance does not define acceleration.
Q2
medium
A ball rolling down a straight slope speeds up uniformly from $3\ \text{m s}^{-1}$ to $15\ \text{m s}^{-1}$ with an acceleration of $4\ \text{m s}^{-2}$. How long does this take?
A$3\ \text{s}$
B$3.75\ \text{s}$
C$4.5\ \text{s}$
D$48\ \text{s}$
Show answer and explanation
Correct answer: A - $3\ \text{s}$
From $v = u + at$, the time is $\frac{15 - 3}{4} = \frac{12}{4} = 3\ \text{s}$. Adding the velocities instead of subtracting them gives $\frac{18}{4} = 4.5\ \text{s}$. Using only the final velocity gives $\frac{15}{4} = 3.75\ \text{s}$, which would be right only if the ball had started from rest, and $48\ \text{s}$ comes from multiplying by the acceleration instead of dividing.
Q3
medium
A merry-go-round carries a child round a circular path of radius $3.5\ \text{m}$, completing one turn every $11\ \text{s}$. Taking $\pi = \frac{22}{7}$, what is the child's average speed?
A$22\ \text{m s}^{-1}$
B$0.32\ \text{m s}^{-1}$
C$1\ \text{m s}^{-1}$
D$2\ \text{m s}^{-1}$
Show answer and explanation
Correct answer: D - $2\ \text{m s}^{-1}$
One turn covers $2\pi R = 2 \times \frac{22}{7} \times 3.5 = 22\ \text{m}$, so the average speed is $\frac{22}{11} = 2\ \text{m s}^{-1}$. Using half the circumference gives $1\ \text{m s}^{-1}$. Dividing the radius by the time gives about $0.32\ \text{m s}^{-1}$, and $22\ \text{m s}^{-1}$ is the distance for one turn that was never divided by the time.
Q4
medium
A lamp post stands $30\ \text{m}$ east of a school gate on a straight road, and a bench stands $50\ \text{m}$ east of the same gate. If the lamp post is now taken as the reference point, what is the position of the bench?
A$20\ \text{m}$ east of the lamp post
B$30\ \text{m}$ east of the lamp post
C$50\ \text{m}$ east of the lamp post
D$80\ \text{m}$ east of the lamp post
Show answer and explanation
Correct answer: A - $20\ \text{m}$ east of the lamp post
Measured from the lamp post, the bench is $50\ \text{m} - 30\ \text{m} = 20\ \text{m}$ to the east. Keeping $50\ \text{m}$ means still measuring from the gate. Adding the two numbers to get $80\ \text{m}$ would only be right if the lamp post lay west of the gate. The value $30\ \text{m}$ is the lamp post's own position measured from the gate.
Q5
easy
A woman walks $700\ \text{m}$ along a straight road from her house to a shop and then walks back to her house along the same road. What are the total distance travelled and the displacement for the whole trip?
Correct answer: C - Distance $1400\ \text{m}$, displacement zero
She walks $700\ \text{m}$ out and $700\ \text{m}$ back, so the path length is $1400\ \text{m}$, while she ends at her house, so the change in position is zero. Swapping the two round is a common slip: a path length can never be zero for a real walk. Counting only the outward leg gives $700\ \text{m}$, and a displacement of $1400\ \text{m}$ would mean she never came home.
Q6
easy
What is meant by uniform circular motion?
AMotion along a straight line at a constant speed
BMotion round a circular path with the speed rising steadily
CMotion round a circular path at a constant speed, with the direction of motion changing all the while
DMotion round a circular path in which both the speed and the direction stay fixed
Show answer and explanation
Correct answer: C - Motion round a circular path at a constant speed, with the direction of motion changing all the while
Uniform circular motion means the object keeps the same speed while going round a circle, so only its direction of motion changes, and it changes at every point. A steadily rising speed would make the motion non-uniform. A constant speed along a straight line is uniform motion but not circular, and a direction that never changed could not bend the path into a circle.
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