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/s2
The acceleration of the car = 8.10 m/s2
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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
v2 = u2 + 2 x a x S
(521 m/s)2 = (0 m/s)2 + 2 x (a) x (0.840 m)
271441 m2/s2 = (0 m/s)2 + (1.68 m) x a
(271441 m2/s2)/(1.68 m) = a
a = 1.62 × 105 m/s2
The acceleration of the bullet = 1.62 × 105 m/s2
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.
The 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/s2
The acceleration of the bus = −1.112 m/s2
Question 4. A particle traveling 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 traveling 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.
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
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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 airplane needs a velocity of 360 km/hr for takeoff. 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 airplane needs a velocity of 360 km/hr for takeoff. 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 =?
∴ v2 = u2 + 2aS
1002 = 02 + 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 the next 60 s. The driver then applies the brakes and brings the train to rest at station Y in the next 5 seconds. Plot a 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 the next 5 s.
Total distance traveled 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
The distance between the two stations X and Y = 1350 m
Question 8. What do you mean by an 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 do distance and displacement vary from each other?
Answer:
Distance and displacement vary from each other:
The shortest distance traveled 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 that has magnitude but no direction is called sa calar 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 seconds. Calculate (1) the distance traveled 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, the diameter of the track, D = 200 m
Therefore, the length of the 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 traveled 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 an athlete completes 3 rounds his displacement is zero
The position of the athlete in the next 1/2 round is just opposite to his starting point
So displacement of the 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 the rate of change of distance with respect to time. The distance traveled 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 an equal distance in equal intervals of time through its motion. An object is said to be moving with non-uniform speed if it covers unequal distances in equal intervals of time.
Question 14. Define uniform and non-uniform velocity.
Answer:
Uniform And Non-Uniform Velocity:
If an object travels an equal distance in an equal interval of time in a specified direction, then the object is said to be moving with uniform velocity. While if an object travels an unequal distance in an 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
∴ The 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 seconds. 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 seconds.
Total distance travelled by the ball = 15 m + 15 m = 30 m
Total time taken = 5 + 2 = 7 s
Average speed = total distance traveled/total time taken
= 30/7
= 4.28 m/s
∴ The average speed of the ball = 4.28 m/s