CBSE SCIENCE
Question: An object covers a distance of 'S' metres in 't' seconds as follows:
Plot a graph, taking 't' on X-axis and 'S' on Y axis.
Answer:
Q.A car travels a certain distance with a speed of 50 km/h and returns with a speed of 40 km/h. Calculate the average speed for the whole journey.
Answer: Let the distance = d km
Total distance travelled by car = 2d
Total time taken 
Average speed
= 44.44 km/h.
Question: An artificial satellite is moving in a circular orbit of radius nearly 42,250 km. Calculate its linear velocity, if it takes 24 hour to revolve round the earth.
Answer: Linear velocity, v
= 3.07 km/s
Question: The given figure shows the position of a body at different times. Calculate
(i) the speed of the body as it moves for 0 to 5 s (ii) 5 to 7 s (iii) 7 to 9 s
Answer: (i) Speed of the body as it moves from 0 to 5 s = Slope of OA
= 0.6 m s-1
(ii) Speed from 5 to 7 s = Slope of AB
= 0 m s-1
(iii) Speed from 7 to 9 s
= 2 m s-1
Question: Plot a distance-time graph for the given data and calculate
(a) the corresponding physical quantity and,
(b) the distance covered by the car at the end of 2.5 s and 6.5 s.
Answer:
The physical quantity, which we get from the distance-time graph, is speed.
To calculate the distance covered at the end of 2.5 s draw a line vertically up from 2.5 s till it meets the distance-time graph. From that point draw a line to the y-axis.
Check where the line meets. The corresponding reading on the y-axis gives the distance travelled. Similarly we can find distance covered at the end of 6.5 s.
The distance covered by the car at the end of 2.5 s = 25 m
The distance covered by the car at the end of 6.5 s = 65 m
Question: What is the nature of the velocity - time graph of a body moving with uniform velocity of 5 m/s.
Answer:
Question: A body moving with a velocity of 40 ms-1 is brought to rest in 5 seconds. Calculate the retardation.
Answer: Given, velocity u = 40m/s v = 0 t = 5 s Retardation = ?
Acceleration
= 
= 
(negative acceleration)
Retardation = 8 m/s2
Question: An object covers a distance of 'S' metres in 't' seconds as follows:
Plot a graph, taking 't' on X-axis and 'S' on Y axis.
Answer:
Answer: Let the distance = d km
Total distance travelled by car = 2d Total time taken Average speed
= 44.44 km/h.Question: An artificial satellite is moving in a circular orbit of radius nearly 42,250 km. Calculate its linear velocity, if it takes 24 hour to revolve round the earth.
Answer: Linear velocity, v
= 3.07 km/sQuestion: The given figure shows the position of a body at different times. Calculate
(i) the speed of the body as it moves for 0 to 5 s (ii) 5 to 7 s (iii) 7 to 9 s
Answer: (i) Speed of the body as it moves from 0 to 5 s = Slope of OA
= 0.6 m s-1(ii) Speed from 5 to 7 s = Slope of AB
= 0 m s-1(iii) Speed from 7 to 9 s
= 2 m s-1Question: Plot a distance-time graph for the given data and calculate
(a) the corresponding physical quantity and,
(b) the distance covered by the car at the end of 2.5 s and 6.5 s.
Answer:
The physical quantity, which we get from the distance-time graph, is speed.
To calculate the distance covered at the end of 2.5 s draw a line vertically up from 2.5 s till it meets the distance-time graph. From that point draw a line to the y-axis.
Check where the line meets. The corresponding reading on the y-axis gives the distance travelled. Similarly we can find distance covered at the end of 6.5 s.
The distance covered by the car at the end of 2.5 s = 25 m
The distance covered by the car at the end of 6.5 s = 65 m
Question: What is the nature of the velocity - time graph of a body moving with uniform velocity of 5 m/s.
Answer:
Question: A body moving with a velocity of 40 ms-1 is brought to rest in 5 seconds. Calculate the retardation.
Answer: Given, velocity u = 40m/s v = 0 t = 5 s Retardation = ?
Acceleration
= = (negative acceleration) Retardation = 8 m/s2
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