Motion
Numerical for practice
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Q1. In a long distance race the athletes were expected to take four rounds of the track such that the line of finish was same as the line of start. Suppose the length of the track was 200 m
(i) What is the total distance to be covered by the athletes?
(ii) What is the displacement of the athletes when they touch the finish line?
(iii) Is the motion of the athletes uniform or non-uniform.
(iv) Is the displacement of an athlete and the distance moved by him at the end of the race equal?
Sol.
(i) Total distance covered = 4 × 200m = 800 m
(ii) As the athletes finish at the starting line
Displacement = final position-initial positive = rA_rA = 0
(iii) Motion is non-uniform as the direction of motion of the athlete is changing while running on the track.
(iv) Displacement and distance moved are not equal.
Q2. A train starting from a railway station and moving with uniform acceleration, attains a speed 40 km/h in 10 minutes. Find its acceleration.
Sol. u = 0 (starting from rest)
v = 40 km/h = 40 × = 11.11 m/s
time t = 10 minutes = 600s
Acceleration (a) = 0.0185 m/s2
Q3. It is estimated that the radio signal takes 1.27 seconds to reach the earth from the surface of the moon. Calculate the distance of the moon from the earth. Speed of radio signal = 3 × 108 ms-1 (speed of light in air).
Sol. Here, time = 1.27 s
speed = 3 × 108 ms-1
distance = ?
Using distance = speed × time, we get
distance = 3 × 108 ms-1 × 1.27 s = 3.81 × 108m = 3.81 = 105 km
Q4. A wireless signal is sent to earth from a spacecraft. This signal reaches the earth in 300 seconds. Calculate the distance of the spacecraft from the earth. Given, speed of the signal = 3 × 108 ms-1.
Sol. Here, t = 300 s
Speed, v = 3 × 108 ms-1
Distance = ?
Using distance = speed × time, we get
distance = 3 × 108 ms_1 × 300 s = 9 × 1010 m = 9 = 107 km
Thus, distance of spacecraft from earth = 9 × 107 km.
Q5. A sound is heard 5 seconds later than the lightning is seen in the sky on a rainy day. Find the distance of the location of lightning. Given speed of sound = 346 ms-1.
Sol. Here, t = 5 s
Speed, v = 346 ms-1
Distance = ?
Using distance = speed × time, we get
distance = 346 ms-1 × 5 s = 1730 m
Thus, distance of the location of lightning = 1730 m
HOPE IT WILL HELP YOU.....
IF SO THAN MARK IT AS BRAINLYT.....
Solved Examples
Q1. In a long distance race the athletes were expected to take four rounds of the track such that the line of finish was same as the line of start. Suppose the length of the track was 200 m
(i) What is the total distance to be covered by the athletes?
(ii) What is the displacement of the athletes when they touch the finish line?
(iii) Is the motion of the athletes uniform or non-uniform.
(iv) Is the displacement of an athlete and the distance moved by him at the end of the race equal?
Sol.
(i) Total distance covered = 4 × 200m = 800 m
(ii) As the athletes finish at the starting line
Displacement = final position-initial positive = rA_rA = 0
(iii) Motion is non-uniform as the direction of motion of the athlete is changing while running on the track.
(iv) Displacement and distance moved are not equal.
Q2. A train starting from a railway station and moving with uniform acceleration, attains a speed 40 km/h in 10 minutes. Find its acceleration.
Sol. u = 0 (starting from rest)
v = 40 km/h = 40 × = 11.11 m/s
time t = 10 minutes = 600s
Acceleration (a) = 0.0185 m/s2
Q3. It is estimated that the radio signal takes 1.27 seconds to reach the earth from the surface of the moon. Calculate the distance of the moon from the earth. Speed of radio signal = 3 × 108 ms-1 (speed of light in air).
Sol. Here, time = 1.27 s
speed = 3 × 108 ms-1
distance = ?
Using distance = speed × time, we get
distance = 3 × 108 ms-1 × 1.27 s = 3.81 × 108m = 3.81 = 105 km
Q4. A wireless signal is sent to earth from a spacecraft. This signal reaches the earth in 300 seconds. Calculate the distance of the spacecraft from the earth. Given, speed of the signal = 3 × 108 ms-1.
Sol. Here, t = 300 s
Speed, v = 3 × 108 ms-1
Distance = ?
Using distance = speed × time, we get
distance = 3 × 108 ms_1 × 300 s = 9 × 1010 m = 9 = 107 km
Thus, distance of spacecraft from earth = 9 × 107 km.
Q5. A sound is heard 5 seconds later than the lightning is seen in the sky on a rainy day. Find the distance of the location of lightning. Given speed of sound = 346 ms-1.
Sol. Here, t = 5 s
Speed, v = 346 ms-1
Distance = ?
Using distance = speed × time, we get
distance = 346 ms-1 × 5 s = 1730 m
Thus, distance of the location of lightning = 1730 m
HOPE IT WILL HELP YOU.....
IF SO THAN MARK IT AS BRAINLYT.....
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