Class 9 physics chapter 1 motion ncert book 15 numerical questions with answers
Answers
Answer:
Numerical questions with answers:
Question:
Bus X travels a distance of 360 km in five hours while bus Y travels a distance of 476 km in 7 hours. Which bus travels faster?
Solution :
For bus X,
Speed= Distance/Time
Speed=360/5=72km/h
For bus Y,
Speed= Distance/Time
Speed=476/7= 68 km/h
The speed of bus X is greater than that of bus Y. Hence, bus X travels faster.
Question:
Arrange the subsequent speeds in growing order (retaining the least pace first):
An athlete jogging with a pace of 10 m/s.
A bicycle shifting with a pace of two hundred m/min.
A scooter shifting with a pace of 30 km/h.
Solution :
Speed of athlete = 10 m/s
Speed of bicycle = two hundred m/min = two hundred/60 m/s = three.33 m/s
Speed of scooter = 30 km/h = 30000/3600 m/s = 8.33 m/s
three.33 m/s < 8.33 m/s < 10 m/s
i.e. 200 m/min < 30 km/h < 10 m/s
QUESTION: A tortoise moves a distance of a hundred meters in 15 minutes. What is the common pace of tortoise in km/h?
Answer:
Total distance = 100m = 0.1 km
Total time taken = 15 minutes = 15/60 = 0.25 hour
Average pace = general distance travelled/general time taken = zero.four km/h
QUESTION. If a sprinter runs a distance of a hundred meters in 9.83 seconds, calculate his common pace in km/h.
Answer:
Total distance travelled = 100m
Total time is taken = 9.83 s
Average pace = general distance travelled/general time taken = 10.172 m/s
Average pace in km/h = 36.62 km/h
QUESTION: A motorcyclist drives from location A to B with a uniform pace of 30 km/h and returns from location B to A with a uniform pace of 20 km/h. Find his common pace.
Answer:
Speed from A to B = 30 km/h
Let D be the gap from A to B
Time is taken, T1 to tour from A to B = distance traveled/pace
T1 = D/30
Speed from B to A = 20 km/h
Time taken, T2 from B to a = distance traveled/pace
T2 = D/20
Total time taken, T = T1 + T2 = D/12
Total distance from A to B and B to A = 2D
Therefore, common pace = general distance travelled/general time taken = 24 km/h
QUESTION: A motorcyclist begins offevolved from relaxation and reaches a pace of 6 m/s after touring with uniform acceleration for 3s. What is his acceleration?
Answer:
Initial velocity = zero m/s
Final velocity = 6 m/s
Time = three sec
a = (v-u)/t
a = 2 m/s^2
QUESTION: A plane touring at 600 km/h hurries up regularly at 10 km/h in keeping with the second. Taking the rate of sound as 1100 km/h at the plane’s altitude how lengthy will it take to attain the ‘sound barrier?
Answer:
Initial velocity, u = 600 km/h
Final velocity, v = 1100 km/h
Acceleration = 10 km/h/s = 600 km/h^2
From formula,
a = (v-u)/t
t = (v-u)/a
t = 50 sec
QUESTION: If a bus touring at 20 m/s is subjected to a regular deceleration of five m/s2, how lengthy will it take to return back to relaxation?
Answer:
Deceleration, a = -five m/s^2
Initial velocity, u = 20 m/s
Final velocity, v = zero m/s
Time, t = ?
a = (v-u)/t
-5 = (zero-20)/t
t = 20/-5 = 4s.
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Answer:
Numerical questions with answers:
Question:
Bus X travels a distance of 360 km in five hours while bus Y travels a distance of 476 km in 7 hours. Which bus travels faster?
Solution :
For bus X,
Speed= Distance/Time
Speed=360/5=72km/h
For bus Y,
Speed= Distance/Time
Speed=476/7= 68 km/h
The speed of bus X is greater than that of bus Y. Hence, bus X travels faster.
Question:
Arrange the subsequent speeds in growing order (retaining the least pace first):
An athlete jogging with a pace of 10 m/s.
A bicycle shifting with a pace of two hundred m/min.
A scooter shifting with a pace of 30 km/h.
Solution :
Speed of athlete = 10 m/s
Speed of bicycle = two hundred m/min = two hundred/60 m/s = three.33 m/s
Speed of scooter = 30 km/h = 30000/3600 m/s = 8.33 m/s
three.33 m/s < 8.33 m/s < 10 m/s
i.e. 200 m/min < 30 km/h < 10 m/s
QUESTION: A tortoise moves a distance of a hundred meters in 15 minutes. What is the common pace of tortoise in km/h?
Answer:
Total distance = 100m = 0.1 km
Total time taken = 15 minutes = 15/60 = 0.25 hour
Average pace = general distance travelled/general time taken = zero.four km/h
QUESTION. If a sprinter runs a distance of a hundred meters in 9.83 seconds, calculate his common pace in km/h.
Answer:
Total distance travelled = 100m
Total time is taken = 9.83 s
Average pace = general distance travelled/general time taken = 10.172 m/s
Average pace in km/h = 36.62 km/h
QUESTION: A motorcyclist drives from location A to B with a uniform pace of 30 km/h and returns from location B to A with a uniform pace of 20 km/h. Find his common pace.
Answer:
Speed from A to B = 30 km/h
Let D be the gap from A to B
Time is taken, T1 to tour from A to B = distance traveled/pace
T1 = D/30
Speed from B to A = 20 km/h
Time taken, T2 from B to a = distance traveled/pace
T2 = D/20
Total time taken, T = T1 + T2 = D/12
Total distance from A to B and B to A = 2D
Therefore, common pace = general distance travelled/general time taken = 24 km/h
QUESTION: A motorcyclist begins offevolved from relaxation and reaches a pace of 6 m/s after touring with uniform acceleration for 3s. What is his acceleration?
Answer:
Initial velocity = zero m/s
Final velocity = 6 m/s
Time = three sec
a = (v-u)/t
a = 2 m/s^2
QUESTION: A plane touring at 600 km/h hurries up regularly at 10 km/h in keeping with the second. Taking the rate of sound as 1100 km/h at the plane’s altitude how lengthy will it take to attain the ‘sound barrier?
Answer:
Initial velocity, u = 600 km/h
Final velocity, v = 1100 km/h
Acceleration = 10 km/h/s = 600 km/h^2
From formula,
a = (v-u)/t
t = (v-u)/a
t = 50 sec
QUESTION: If a bus touring at 20 m/s is subjected to a regular deceleration of five m/s2, how lengthy will it take to return back to relaxation?
Answer:
Deceleration, a = -five m/s^2
Initial velocity, u = 20 m/s
Final velocity, v = zero m/s
Time, t = ?
a = (v-u)/t
-5 = (zero-20)/t
t = 20/-5 = 4s.
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