Class 9 Describing Motion Around Us Mock Test (CBSE)

Topics in Describing Motion Around Us

• Describing position
• Distance and displacement
• Average speed and average velocity
• Average acceleration
• Position-time graphs
• Velocity-time graphs
• Kinematic equations for constant acceleration
• Motion in a plane
• Uniform circular motion

Describing Motion Around Us practice questions with answers

1. Q1Multiple choice·1 mark
   The area under a velocity-time graph represents:

   1. (a)Velocity
   2. (b)Displacement
   3. (c)Acceleration
   4. (d)Distance-time ratio
   Show answer and step marks

   1. Correct option: (b) Displacement (Area under velocity-time graph gives product of velocity and time, which is displacement)+1

   Answer: (b)
2. Q2Multiple choice·1 mark
   An athlete completes one round of a circular track of diameter $d$ in $40\text{ s}$. What will be the displacement at the end of $2\text{ min } 20\text{ s}$?

   1. (a)Zero
   2. (b)$d$
   3. (c)$2d$
   4. (d)$7\pi d$
   Show answer and step marks

   1. Correct option: (b) $d$ (Total time = $140\text{ s}$, number of rounds = $3.5$, displacement after $3.5$ rounds is equal to the diameter $d$)+1

   Answer: (b)
3. Q3Multiple choice·1 mark
   The slope of a distance-time graph for the motion of an object represents its:

   1. (a)Acceleration
   2. (b)Speed
   3. (c)Displacement
   4. (d)Distance
   Show answer and step marks

   1. Correct option: (b) Speed (Slope of distance-time graph $\frac{\Delta s}{\Delta t}$ gives speed)+1

   Answer: (b)
4. Q4Multiple choice·1 mark
   A car accelerates uniformly from $18\text{ km/h}$ to $36\text{ km/h}$ in $5\text{ s}$. The acceleration of the car is:

   1. (a)$1\text{ m/s}^2$
   2. (b)$2\text{ m/s}^2$
   3. (c)$3.6\text{ m/s}^2$
   4. (d)$5\text{ m/s}^2$
   Show answer and step marks

   1. Correct option: (a) $1\text{ m/s}^2$ ($u = 5\text{ m/s}$, $v = 10\text{ m/s}$, $a = \frac{10 - 5}{5} = 1\text{ m/s}^2$)+1

   Answer: (a)
5. Q5Short answer·3 marks
   Ravi is cleaning his kitchen. He walks $3\text{ m}$ in a straight line from the refrigerator to the sink. He then walks $4\text{ m}$ in a straight line from the sink to the dining table, along a path at right angles to his first walk. (a) Find the total distance Ravi walks from the refrigerator to the dining table. (b) Find the magnitude of his displacement from the refrigerator to the dining table. (c) Explain why the displacement is not equal to the distance walked.
   Show answer and step marks

   1. Distance walked = $3\text{ m} + 4\text{ m} = 7\text{ m}$+1
   2. Displacement is the straight line from refrigerator to table, a hypotenuse: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5\text{ m}$+1
   3. Displacement is the shortest straight-line distance between the start and end points, whereas distance counts the full path length, so the displacement ($5\text{ m}$) is less than the distance ($7\text{ m}$) because Ravi's path is not a straight line+1

   Answer: (a) Distance = $7\text{ m}$; (b) Displacement = $5\text{ m}$; (c) Displacement is the straight-line shortest separation of the end points, while distance is the full path length, and the path here is bent at a right angle.
6. Q6Short answer·3 marks
   A delivery drone leaves a warehouse and flies $12\text{ km}$ due north, then turns and flies $5\text{ km}$ due east. The whole trip takes $26\text{ minutes}$. Find the magnitude of the drone's average velocity for the whole trip, in km/h.
   Show answer and step marks

   1. Displacement is the straight line from start to end, perpendicular components combine: $\sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ km}$+1
   2. Time taken $= \dfrac{26}{60}\text{ h} = \dfrac{13}{30}\text{ h}$+1
   3. Average velocity $= \dfrac{13\text{ km}}{\frac{13}{30}\text{ h}} = 30\text{ km/h}$+1

   Answer: $30\text{ km/h}$

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