Motion

Exercise Question Answer

Part 2

Question 6: Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

Which of the three is travelling the fastest?
Are all three ever at the same point on the road?
How far has C travelled when B passes A?
How far has B travelled by the time it passes C?

Answer:

(a) It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.
(b) All of them never come at the same point at the same time.
(c) According to graph; each small division shows about 0.57 km.

A is passing B at point S which is in line with point P (on the distance axis) and shows about 9.14 km

Thus, at this point C travels about
9.14 – (0.57 x 3.75) km = 9.14 km – 2.1375 km = 7.0025 km ≈ 7 km
Thus, when A passes B, C travels about 7 km.

(d) B passes C at point Q at the distance axis which is ≈ 4km + 0.57km x 2.25 = 5.28 km

Therefore, B travelled about 5.28 km when passes to C.

Question 7: A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s², with what velocity will it strike the ground? After what time will it strike the ground?

Answer: Here we have, Initial velocity,u=0
Distance,s=20m
Acceleration,a= 10 m s^-2
Final velocity,v=?
Time,t= ?

(a) Calculation of Final velocity, `v`

We know that, `v^2= u^2 + 2as`

`=> v^2 = 0 + 2 xx 10\ m//s^2 xx 20m`

`=> v^2 = 400\ m^2s^(-2)`

`=> v = sqrt(400\ m^2 s^(-2))`

`=> v = 20\ m s^(-1)`

(b) Calculation of time, `t`

We know that, `v=u+at`

`=> 20\ ms^(-1) = 0+10\ m s^(-2) xx t`

`=> t = (20\ ms^(-1))/(10\ ms^(-2)) = 2s`

Therefore, Ball will strike the ground at a velocity of `20\ ms^(-1)`

Time taken to reach at the ground `= 2s`

Question 8: The speed-time graph for a car is shown is Fig. 8.12.

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?

Answer:

(a) Distance travelled by car in the 4 second

The area under the slope of the speed – time graph gives the distance travelled by an object.

In the given graph

56 full squares and 12 half squares come under the area slope for the time of 4 second.

Total number of squares = 56 + 12/2 = 62 squares

The total area of the squares will give the distance travelled by the car in 4 second.

On the time axis, 5 squares = 2s

∴ 1 square `=2/5s`

On the speed asix 3 squares = 2 m/s

∴ 1 square `= 2/3 m//s`

∴ area of 1 square `=2/5 s xx 2/3 m//s = 4/15 m`

∴ area of 62 squares `= 4/15 m xx 62`

`248/15 m = 16.53 m`

Therefore, car travels 16.53 m in first 4 second.

(b) Part MN of the slope of the graph is straight line parallel to the time axis, thus this potion of graph represents uniform motion of car.

Question 9: State which of the following situations are possible and give an example for each of these:

(a) An object with a constant acceleration but with zero velocity

Answer: When an object is thrown upwards, a time comes when its velocity becomes zero. Even at that point, acceleration due to gravity is working on it. This is an example of object under uniform acceleration with zero velocity.

(b) An object moving in a certain direction with an acceleration in the perpendicular direction.

Answer: When an object is moving with uniform circular motion, its path and acceleration are in mutually perpendicular direction

Question 10: An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Answer: Here we have, Radius, r = 42250km
Time, t = 24 hours
Speed =?

We know that velocity along a circular path `= (2pir)/text{time}`

`=> v = (2 xx 22/7 xx 42250\ km)/(24h)`

`=>v = (2xx22 xx42250)/(7xx24)\ km//h`

` =>v= 11065.47\ km//h`

Thus, speed of the given satellite `= 11065.47\ km//h`