Grade 9 – Physics – Motion – Important Question and Answers

    Question 1. State the meaning of uniform circular motion.
    Answer. When a body moves with a uniform speed in a circular path, the body is said to describe uniform circular motion.

    Question 2. State the type of motion represented by the given graph.

    Answer: Uniform motion.

    Question 3: What does the slope of a velocity-time graph represent?

    Answer. Acceleration.

    Question 4. If the acceleration of the particle is constant in magnitude but not in direction, what type of path does the particle follow?
    Answer. Circular path.

    Question 5: What kind of motion of a body is represented by the graphs given below?


    (a) Stationary
    (b) The body is moving with a uniform speed for sometime then stops.

    Question 6: Name the quantity measured by the area occupied below the velocity-time graph.
    Answer. Displacement.

    Question 7: Define uniform motion.
    Answer. When a body covers equal distances in equal intervals of time, however small may be time intervals, the body is said to describe a uniform motion.

    Question 8:. Under what condition is the magnitude of distance and displacement equal?
    Answer. When a body is moving in a linear path without changing its direction.

    Question 9: An object starts with initial velocity u and attains a final velocity of v. The velocity of the object is changing at a uniform rate.Write the formula for calculating the average velocity $V_{av}$.

    Answer: \[V_{av}=\frac{u+v}{2}\]

    Question 10. A body is thrown in the vertically upward direction rises upto a height ‘h’ and comes back to the position of start. Calculate
    (a) the total distance travelled by the body (b) the displacement of the body
    Answer. (a) Total distance = h + h = 2h
    (b) The displacement = 0 [$\because$ Body comes back to the position of start]

    Question 11. Give the name of the physical quantity that corresponds to the rate of change of motion and state its unit in SI system.
    Answer. The physical quantity is speed. The unit of speed in SI system is $ms^{-1}$.

    Question 12. Which amongst the following is a vector quantity and why? (i) Distance (ii) Displacement.
    Answer. Displacement is a vector quantity. It is because the direction is specified in case of displacement.

    Question 13. A cyclist travels a distance of 4 km from P to Q and then moves a distance of 3 km at right angle to PQ. Find his resultant displacement graphically.

    Answer. The distance covered by the cyclist is shown graphically.
    From the figure, resultant displacement,

    $\mathrm{PR}=\sqrt{\mathrm{PQ}^{2}+\mathrm{QR}^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{25}=5 \mathrm{km}$

    Question 14. A bus decreases its speed from $80 \mathrm{km} / \mathrm{h}$ to $50 \mathrm{km} / \mathrm{h}$ in $4 \mathrm{s}$. Find the acceleration of the bus.
    Answer. \[u=80 \mathrm{km} / \mathrm{h}=80 \times \frac{5}{18} \mathrm{m} / \mathrm{s}=\frac{200}{9} \mathrm{m} / \mathrm{s}\]\[v=50 \mathrm{km} / \mathrm{h}=50 \times \frac{5}{18} \mathrm{m} / \mathrm{s}=\frac{125}{9} \mathrm{m} / \mathrm{s}\]\[t=4 \mathrm{s}\] \[\mathrm{Acceleration,} \ a=\frac{v-u}{t}=\frac{1}{4} (\frac{125}{9}-\frac{200}{9})=\frac{-75}{9 \times 4}=-2.08 \mathrm{m} / \mathrm{s}^{2}\]

    Question 15. An electric train is moving with a velocity of $120 \mathrm{km} / \mathrm{hr}$. How much distance will it cover in $30 \mathrm{s} ?$
    Answer. \[\mathrm{Velocity}\ v=120 \mathrm{km} / \mathrm{hr}=120 \times \frac{5}{18} \mathrm{m} / \mathrm{s}=\frac{100}{3} \mathrm{m} / \mathrm{s}\]
    \[\mathrm{Time}\ t=30 \mathrm{s}\]
    \[\mathrm{Required\ covered\ distance}\ =v t=\frac{100}{3} \mathrm{m} / \mathrm{s} \times 30 \mathrm{s}=1000 \mathrm{m}=1 \mathrm{km}\]

    Question 16. Represent the given data graphically.

    Answer :

    Question 17: Explain the differences between the two graphs.

    Answer. Both are uniform accelerated motions, but in first case initial velocity is zero while in second case it has some fixed value. Also in first case rate of change of velocity i.e., acceleration is greater than in the second case.

    Question 18:. A body can have zero average velocity but not zero average speed. Justify.
    Answer. Displacement can have zero value but distance has always some fixed value for the moving object. When a body after covering some distance, it comes back to his initial position, then displacement has zero value but distance has some fixed value, because distance is the length of path covered. Thus, in that case,

    Average velocity $=\frac{\text { displacement }}{\text { time taken }}=\frac{0}{t}=0$
    But average speed $=\frac{\text { Total distance }}{\text { Time taken }}=$ A fixed value.

    Question 19:. A train 100 m long is moving with a velocity of $60 \mathrm{kmh}^{-1} .$ Find the time it takes to cross the bridge $1 \mathrm{km}$ long.

    Answer. Velocity, $v=60 \mathrm{kmh}^{-1}=60 \times \frac{5}{18} \mathrm{m} / \mathrm{s}=\frac{50}{3} \mathrm{m} / \mathrm{s}$
    Distance covered $=100 \mathrm{m}+1 \mathrm{km}=100 \mathrm{m}+1000 \mathrm{m}=1100 \mathrm{m}$
    Time taken $=\frac{\text { Distance covered }}{\text { Velocity }}=\frac{1100}{50 / 3}=\frac{1100 \times 3}{50}=66 \mathrm{s}$

    Question 20:What do the graphs shown below indicate?

    Answer. In first graph, speed is decreasing uniformly and after some time it becomes to zero. But in second case firstly speed is decreasing uniformly and after some time it starts to increasing uniformly.

    Question 21: A marble rolling on a smooth floor has an initial velocity of $0.4 \mathrm{m} / \mathrm{s}$. If the floor offers a retardation of $0.02 \mathrm{m} / \mathrm{s}^{2},$ calculate the time it will take to come to rest.

    Answer. $u=0.4 \mathrm{m} / \mathrm{s}, a=-0.02 \mathrm{m} / \mathrm{s}^{2}, v=0, t=?$
    Time, $t=\frac{v-u}{a}=\frac{0-0.4}{-0.02}=\frac{0.4}{0.02}=20 \mathrm{s}$

    Question 22:An athlete completes one round of a circular track of diameter 49 m in 20 s. Calculate the distance covered and displacement at the end of 30 s.


    In 20 s athlete completes 1 round.
    So, in 30 s athlete will complete $\frac{1}{20} \times 30=\frac{3}{2}$ round.

    If athlete starts his round from point A, then after completing his $\frac{3}{2}$ round he will reach at point ${B}$. Now, circumference of the circular track $={p} \times {D}=\frac{22}{7} \cdot 49=154$ ${ m}$

    Thus, distance covered in $\frac{3}{2}$ round $=\frac{3}{2} \times 154$ m$=231$ $ {m}$

    Displacement $={AB}=$ Diameter $=49$ ${ m}$

    Question 23. What can you say about the motion of an object whose distance-time graph is :
    (i) a straight line, parallel to the time axis.
    (ii) a straight line passing through the origin making an angle from the time axis.


    (i) In distance-time graph, a straight line parallel to time axis indicates that with increase of time interval, there is no any change in distance so the object is stationary.
    (ii) In this case body is covered equal distance in equal interval of time, so the body is moving with uniform speed starting from rest.

    Question 24. Derive the equation $v^2 – u^2 = 2as $ graphically.


    Figure alongside represents a velocity-time graph BC, in which AB represents the initial velocity u, CE represents the final velocity v, such that change in velocity is represented by CD, which takes place in time
    t, represented by AE.
    Now, distance covered = Area of trapezium ABCE

    $s=\frac{1}{2}(A B+C E) \times A E$
    $\Rightarrow s=\frac{1}{2}(u+v) \times t$
    $\Rightarrow s=\frac{1}{2}(v+u) \times \frac{v-u}{a}$ [$\because t=\frac{v-u}{a}$]

    $\Rightarrow 2as = v ^2 – u ^2$ $ \Rightarrow v ^2 – u ^2 = 2as$

    Question 25.

    (a) A car accelerates uniformly from $18 \mathrm{kmh}^{-1}$ to $36 \mathrm{km} / \mathrm{h}^{-1}$ in 5 s. Calculate :
    (i) acceleration
    (ii) distance covered by the car in that time
    (b) The length of minute hand of a clock is $14 \mathrm{cm} .$ Calculate the speed with which the tip of the minute hand moves.


    (a) $u=15 \ \mathrm{kmh}^{-1}=18 \times \frac{5}{18} \ \mathrm{ms}^{-1}=5 \ \mathrm{ms}^{-1}$
    $v=36 \ \mathrm{kmh}^{-1}=36 \times \frac{5}{18} \ \mathrm{ms}^{-1}=10 \ \mathrm{ms}^{-1}, t=5 \ \mathrm{s}$

    (i) Acceleration, $a=\frac{v-u}{t}=\frac{10 \ \mathrm{ms}^{-1}-5 \ \mathrm{ms}^{-1}}{5\ \mathrm{s}}=\frac{5 \ \mathrm{ms}^{-1}}{5 \ \mathrm{s}}=1 \ \mathrm{ms}^{-2}$
    (ii) Distance covered, $s=u t+\frac{1}{2} a t^{2}=5 \times 5+\frac{1}{2} \times 1 \times 5^{2}=25+12.5=37.5 \ \mathrm{m}$

    (b) Circumference of the clock $=2 \pi r=2 \times \frac{22}{7} \times 14=88\ \mathrm{cm}$
    Time taken by minute hand in one complete rotation $=60 \ \mathrm{min}=60 \times 60 \ \mathrm{s}=3600 \ \mathrm{s}$
    Speed of the tip of the minute hand $=\frac{\text { Distance covered }}{\text { Time taken }}=\frac{88 \ \mathrm{cm}}{3600 \ \mathrm{s}}=0.024 \ \mathrm{cm} / \ \mathrm{s}$