## Chapter 1 Motion Short Answer Type Question And Answers

**Question 1. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.**

**Answer. **

**Given:**

A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m.

Distance S = 110 m

Time t = 5.21 s

Initial Velocity u = 0

S = u × t + 0.5 × a × t2 = 110

= 0 × 5.21 + 0.5 a × 5.21 × 5.21

a = \(\frac{110}{13.57}\)

= 8.10 m/s^{2}

**Question 2. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).**

**Answer**.

**Given:**

A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m.

Initial velocity u = 0 m/s

Final velocity v = 521 m/s

Distance S = 0.840 m

v^{2} = u^{2} + 2 x a x S

(521 m/s)^{2} = (0 m/s)^{2} + 2 x (a) x (0.840 m)

271441 m^{2}/s^{2} = (0 m/s)^{2} + (1.68 m) x a

(271441 m^{2}/s^{2})/(1.68 m) = a

a = 1.62 × 105 m/s^{2}

**Question 3. A bus decreases its speed from 80 km h−1 to 60 kmh−1 in 5 s. Find the acceleration of the bus.**

**Answer. **

**Given:**

A bus decreases its speed from 80 km h−1 to 60 kmh−1 in 5 s.

Initial speed of bus u = 80 km/hr

= 80 × 5/18 = 22.22 m/s.

Final speed of bus v = 60 km/hr

= 60 × 5/18 = 16.66 m/s.

Time taken to decrease the speed t = 5 s

Acceleration a = (v – u )/t

= 16.66 – 22.22/5

= −1.112 m/s^{2}

**Question 4. A particle travelling along a straight line with a constant speed covers distance AB = 600 m in 20 s and returns to the starting point A in another 30 s. Calculate its**

**Average speed and****Average velocity**

**Answer.**

**Given:**

A particle travelling along a straight line with a constant speed covers distance AB = 600 m in 20 s and returns to the starting point A in another 30 s.

\(\text { Average speed }=\frac{\text { Total distance travelled }}{\text { Time taken }} \)= \(\frac{(600 m+600 m)}{(20 s+30 s)}\)

= \(\frac{1200 \mathrm{~m}}{50 \mathrm{~s}}\)

= 24ms^{-1}

**Question 5. An electron describes a semicircular path of radius 14 cm is 4.4 µs. Calculate its**

**Speed****Velocity**

**Answer. **

**Given:**

An electron describes a semicircular path of radius 14 cm is 4.4 µs.

(1) \(\text { Speed }=\frac{\text { Distance travelled }}{\text { Time taken }}\)

i.e., \(v=\frac{\text { Length of semicircle }}{\text { Time }}\)

i.e., v = \(\frac{\pi r}{t}\)

v = \(\frac{22}{7} \times \frac{14 \times 10^{-2} \mathrm{~m}}{4.4 \times 10^{-6} \mathrm{~s}}\)

= \(\frac{44}{4.4} \times 10^4 \mathrm{~ms}^{-1}\)

= 1 × 105 ms^{-1}

(2) \(\text { Velocity }=\frac{\text { Displacement }}{\text { Time }}\)

\(=\frac{\text { Diameter of the circle }}{\text { Time }}\)= \(\frac{2 r}{t}\)

= \(\frac{2 \times 14 \times 10^{-2} \mathrm{~m}}{4.4 \times 10^{-6} \mathrm{~s}}\)

= \(\frac{280}{44} \times 10^4 \mathrm{~ms}^{-1}\)

= 6.38 × 104 ms^{-1}

**Question 6. An aeroplane needs a velocity of 360 km/hr for take off. If it starts from rest and accelerates at the rate of 4 ms−2, what must be the minimum length of the runway?**

**Answer. **

**Given:**

An aeroplane needs a velocity of 360 km/hr for take off. If it starts from rest and accelerates at the rate of 4 ms−2,

Given u = 0

v = 360 km/hr

= \(360 \times \frac{1000}{3600}=100 \mathrm{~ms}^{-1}\)

a = 4 ms^{-2}

S = ?

∴ v^{2} = u^{2} + 2aS

100^{2} = 0^{2} + 2 × 4 × S

100 × 100 = 8 S

∴ \(S=\frac{100 \times 100}{8}=1250 \mathrm{~m}\)

So, the minimum length of the runway = 1250 m

**Question 7. A metro train starts from station X and accelerates uniformly at the rate of 2 ms ^{-2} for 10 s. Then, it maintains its speed for next 60 s. The driver then applies the brakes and brings the train to rest at station Y in next 5 s. Plot velocity-time graph of the Metro train and find the distance between the two stations X and Y.**

**Answer.**

**Given:**

A metro train starts from station X and accelerates uniformly at the rate of 2 ms^{-2} for 10 s. Then, it maintains its speed for next 60 s.

The driver then applies the brakes and brings the train to rest at station Y in next 5 s.

Total distance travelled by the metro train = Area of trapezium ABCD

= \(\frac{1}{2} B M \times(A D+B C)\)

= \(\frac{1}{2} \times 20(75+60)\)

= 10 × 135

= 1350 m

**Question 8. What do you mean by object at rest?**

**Answer:**

**Object at rest:**

If an object does not change its position with respect to its surroundings then it is said to be at rest.

**Question 9. How distance and displacement vary from each other?**

**Answer:**

**Distance and displacement vary from each other:**

The shortest distance travelled by an object in a definite direction is termed as displacement. On the other hand, distance is the total length of the path covered by an object, irrespective of the direction it takes.

**Question 10. Differentiate between scalar and vector quantity.**

**Answer:**

**Difference between scalar and vector quantity:**

The physical quantity which has magnitude but no direction is called scalar quantity. Vector quantity, on the other hand, is a physical quantity which has both magnitude and direction.

**Question 11. An athlete completes a round of a circular track of diameter 200 m in 20 s. Calculate (1) the distance travelled by the athlete (2) the magnitude of the displacement of the athlete at the end of 1 minute and 10 seconds. **

**Answer: **

**Given:**

An athlete completes a round of a circular track of diameter 200 m in 20 s.

Here, diameter of the track, D = 200 m

Therefore length of circular track = circumference of the circular track

= 2πr = π (2r) = πD

= \(\frac{22}{7} \times 200=628.57 \mathrm{~m}\)

(1) Distance travelled in 20 s = length of circular track = 628.57 m

Distance travelled in 1 s = \(\frac{628.57}{20} \mathrm{~m}=31.43 \mathrm{~m}\)

Distance travelled in 1 minute and 10 s (or 70 s)

= \(\frac{628.57}{20} \mathrm{~m} \times 70\)

= 2199.99 = 2200 m

(2) Number of rounds completed in 20 s = 1

Number of rounds completed in 70 s

= \(\frac{1}{20} \times 70\)

= \(3 \frac{1}{2}\)

When athlete completes 3 rounds his displacement is zero

The position of the athlete in next 1/2 round is just opposite to his starting point

So displacement of athlete at the end of 1 minute and 10 s = diameter of the circular track = 200 m

**Question 12. Define speed. Write its SI unit.**

**Answer:**

**Speed:**

Speed is defined as rate of change of distance with respect to time. The distance travelled by a body in unit time interval is called speed. Its SI unit is metre/second (m/s).

**Question 13. Differentiate between uniform and non-uniform speed.**

**Answer:**

**Difference between uniform and non-uniform speed:**

An object is said to be at uniform speed if it travels equal distance in equal interval of time through its motion. An object is said to be moving with non-uniform speed if it covers unequal distances in equal interval of time.

**Question 14. Define uniform and non-uniform velocity.**

**Answer:**

**Uniform and non-uniform velocity:**

If an object travels equal distance in equal interval of time in a specified direction, then the object is said to be moving with uniform velocity. While if an object travels unequal distance in equal interval of time in a specified direction, then the object is said to be moving with non-uniform velocity.

**Question 15. A car travels a distance of 3 km in 5 minutes. Calculate the speed in cm/s.**

**Answer:**

**Given:**

A car travels a distance of 3 km in 5 minutes.

Distance = 3 km

As we know 1 km = 1000 m, 1 m = 100 cm

Therefore, 3 km = 3 × 1000 × 100 = 300000 cm

Time = 5 minutes

1 min = 60 sec

Therefore, 5 min = 5 × 60 = 300 s

\(\text { Speed }=\frac{\text { distance }}{\text { time }}\)Speed = 300000/300 = 1000 cm/s

**Answer:** Speed of the car is 1000 cm/s

**Question 16. A ball travels 15 m in 5 seconds then another 15 m distance in 2 second. What is the average speed of the ball?**

**Answer:**

**Given:**

A ball travels 15 m in 5 seconds then another 15 m distance in 2 second.

Total distance travelled by the ball = 15 m + 15 m = 30 m

Total time taken = 5 + 2 = 7 s

Average speed = total distance travelled/total time taken

= 30/7

= 4.28 m/s