**Page 100
Q.1 Â Â An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
**

Â Â Â Â Â Â Suppose a ball starts moving from point A and it returns back at same point A, then the distance will be equal to 20 meters while displacement will be zero.

**Q.2 Â Â A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
**

Â Â Â Â Â Â Given, side of the square field = 10m

Â Â Â Â Â Â Therefore, perimeter = 10 m x 4 = 40 m

Â Â Â Â Â Â Farmer moves along the boundary in 40s.

Â Â Â Â Â Â Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?

Â Â Â Â Â Â Since in 40 s farmer moves 40 mTherefore, in 1s distance covered by farmer =Â Â

Â Â Â Â Â Â Therefore, in 140s distance covered by farmer = 1 Ã— 140 m = 140 m

Â Â Â Â Â Â Now, number of rotation to cover 140 along the boundry Â = Â

Â Â Â Â Â Â

Â Â Â Â Â Â Thus after 3.5 round farmer will at point C of the field.

Â Â Â Â Â Â Therefore, DisplacmentÂ

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Thus, after 2 minute 20 second the displacement of farmer will be equal toÂ north east from initial position.Â

**Q.3 Â Â Which of the following is true for displacement?
**

**Page 102**

**Q.1 Â Â Distinguish between speed and velocity.
**

**Q.2 Â Â Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
**

**Q.3 Â Â What does the odometer of an automobile measure?
**

**Q.4 Â Â What does the path of an object look like when it is in uniform motion?
**

**Q.5 Â Â During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, ****.
**

Â Â Â Â Â Time = 5 minute =Â second

Â Â Â Â Â We know that, Distance = Speed Ã— Time

Â Â Â Â Â Distance =Â

**Page 103**

**Q.1 Â Â When will you say a body is in
Â Â Â Â Â Â (i) uniform acceleration?
**

Â Â Â Â Â Â (ii) A body is said in non-uniform acceleration when its motion is along a straight line and its velocity changes by unequal magnitude in equal interval of time.

**Q.2 Â Â A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.
**

Â Â Â Â Â Â Therefore, acceleration, a = ?

Â Â Â Â Â Â We know that, v = u + at

Â Â Â Â Â Â 60 km/ h = 80km/h + a Ã— 5s

Â Â Â Â Â Â 60km/h â€“ 80km/h = a Ã— 5s

Â Â Â Â Â Â â€“ 20km/h = a Ã— 5s

Â Â Â Â Â Â a =Â

Â Â Â Â Â Â a = â€“ 4km/h/s

Â Â Â Â Â Â Therefore, Acceleration = â€“4 km/h/s or, â€“1.1Â Â Â

**Q. 3 Â Â A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.
**

Â Â Â Â Â Â Initial velocity, u = 0,

Â Â Â Â Â Â Final velocity, v = 40km/h = 11.11m/s

Â Â Â Â Â Â Time (t) = 10 minute = 60 x 10=600s

Â Â Â Â Â Â Acceleration (a) =?

Â Â Â Â Â Â We know that, v = u + at

Â Â Â Â Â Â 40km/h = 0km/h + a Ã— 10m

Â Â Â Â Â Â 11.11 m/s = a Ã— 600s

Â Â Â Â Â Â a =Â

**Page 107**

**Q.1 Â Â What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?** * Sol. Â Â Â *(a) The slope of the distance-time graph for an object in uniform motion is straight line.

Â Â Â Â Â Â (b) The slope of the distance-time graph for an object in non-uniform motion is not a straight line.

**Q.2 Â Â What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
**

**Q.3 Â Â What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
**

**Q.4 Â Â What is the quantity which is measured by the area occupied below the velocity-time graph?
**

**Page 109**

**Q.1 Â Â A bus starting from rest moves with a uniform acceleration of 0.1 m s ^{-2}Â for 2 minutes. Find (a)the speed acquired, (b) the distance travelled.
**

Â Â Â Â Â Â Initial velocity (u) = 0

Â Â Â Â Â Â Acceleration (a) = 0.1ms-2

Â Â Â Â Â Â Time (t) = 2 minute = 120 second

Â Â Â Â Â Â (a) The speed acquired:

Â Â Â Â Â Â We know that, v = u + at

Â Â Â Â Â Â â‡’ v = 0 + 0.1m/s2 x 120 s

Â Â Â Â Â Â â‡’ v = 120 m/s

Â Â Â Â Â Â Thus, the bus will acquire a speed of 120 m/s after 2 minute with the given acceleration.

Â Â Â Â Â Â (b) The distance travelled:

Â Â Â Â Â Â We know that,

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Thus, bus will tavel a distance of 7200 m or 7.2 km in the given time of 2 minute.Â Â

**Page 110**

**Q.2 Â Â A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of â€“ 0.5 m s ^{-2}. Find how far the train will go before it is brought to rest.
**

Â Â Â Â Â Â Initial velocity, u = 90km/h

Â Â Â Â Â Â

Â Â Â Â Â Â Final velocity, v = 0

Â Â Â Â Â Â Acceleration,a = â€“ 0.5Â

Â Â Â Â Â Â Therefore, distance travelled = ?

Â Â Â Â Â Â We know tha,Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Therefore, train will go 625 m be fore it brought to rest.

**Q.3 Â Â A trolley, while going down an inclined plane, has an acceleration of 2 cm s ^{-2}. What will be its velocity 3 s after the start?
**

Â Â Â Â Â Â Initial velocity, u = 0

Â Â Â Â Â Â Acceleration (a) = 2cm/s

Â Â Â Â Â Â Therefore, Final velocity, v = ?

Â Â Â Â Â Â We know that, v = u + at

Â Â Â Â Â Â Therefore, v = 0 + 0.02Ã— 3s

Â Â Â Â Â Â v = 0.06 m/s

Â Â Â Â Â Â Therefore the final velocity of trolley will be 0.06 m/s after start.Â

**Q.4 Â Â A racing car has a uniform acceleration of 4 m s ^{-2}. What distance will it cover in 10 s after start?**

Â Â Â Â Â Â Acceleration, a = 4m/s2

Â Â Â Â Â Â Initial velocity, u = 0

Â Â Â Â Â Â Time, t = 10s

Â Â Â Â Â Â Therefore, Distance (s) covered =?

Â Â Â Â Â Â We know that, Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Thus, racing car will cover a distance of 200m after start in 10 s with given acceleration.

**Q. 5 Â Â A stone is thrown in a vertically upward direction with a velocity of 5 m s ^{-1}. If the acceleration of the stone during its motion is 10 m s^{-2}Â in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?**

Â Â Â Â Â Â Initial velocity (u) = 5m/s

Â Â Â Â Â Â Â Final velocity (v) = 0 (Since from where stone starts falling its velocity will become zero)

Â Â Â Â Â Â Acceleration (a) = â€“ 10m/s

Â Â Â Â Â Â Height, i.e. Distance, s =?

Â Â Â Â Â Â Time (t) taken to reach the height =?

Â Â Â Â Â Â We know that,Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Now, we know that , v = u + at

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Thus, stone will attain a height of 1.25m And time taken to attain the height is 0.5s Â Â

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sorry but regarding

Q.5 During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3Ã—108msâˆ’1.

Sol. Here we have, speed = 3Ã—108msâˆ’1

Time = 5 minute = 5Ã—60s=300 second

We know that, Distance = Speed Ã— Time

â‡’ Distance = 3Ã—108msâˆ’1Ã—300s=1800Ã—108m=1.8Ã—1011m

but 3*300=900 or the above please clarify.

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