The document NCERT Exemplar - Motion Class 9 Notes | EduRev is a part of the Class 9 Course Science Class 9.

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**Short Answer Type Questions**

**Q.1. The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify your answer.****Ans. **No, the object might have travelled some distance, provided the initial position and final position of the body coincide.

Example of zero displacement**Q.2. How will the equations of motion for an object moving with a uniform velocity change?****Ans. **When an object travels with uniform velocity then v = u and a = 0.

∴ v = u + at ⇒ v = u

► s = ut + (1/2) at^{2} = ut

► v^{2} = u^{2} + 2as ⇒ v^{2} = u^{2}**Q.3. A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in the figure. Plot a velocity-time graph for the same.****Ans:****Q.4. A car starts from rest and moves along the jc-axis with constant acceleration of 5 ms ^{-}^{2} for 8 seconds. It then continues with constant velocity. What distance will the car cover in 12 seconds since it started from rest?**

= 160 m (for first 8 s)

= 40 ms

∴ s

= 320 m

Let distance travelled from A to B = S km

∴ Total distance = x + x = 2x.

► Time taken to travel from A to B = t

► Time taken to travel from B to A = t

► Total time,

► Average speed =

Ans.

**Long Answer Type Questions**

**Q.7. An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with the same acceleration? How does the difference in heights vary with time?****Ans.**

__Initially difference between height of two objects:__= 150 - 100 = 50 m

__Distance travelled by first obejct in 2 s:__

__Distance travelled by second object in 2 seconds:__

__Height of first object at t = 2s:__

= 150 - 20 = 130 m

__Height of second object at t = 2s:__

= 100 - 20 = 80 m

__Difference in height after t = 2s:__

= 130 - 80 = 50 m**Q.8. An object starting from rest travels 20 m in first 2s and 160 m in next 4s. What will be the velocity after 7s from the start?****Ans. **u = 0, s_{1} = 20 m, s_{2} = 160 m

u = 0, s = 20 m, t = 2 s

s = ut + at^{2}

or a = 2 s/t^{2} = 10 ms^{-2}

v = u + at = 10 x 2 = 20 ms^{-}^{1}.__For second interval:__u = 20 ms

Again, s = ut + at

160 = 20 x 4 + x a x 16a = 10 ms

i.e. 'a' is constant

= 70 m/s

v

= =1 m/sv

v

u = 5 x 10

► 2u = u + 10

► u x 10

► 5 x 10

or t = 5 s.

= 5 x 10

= 3.75 x 10

If u = initial velocity, then final velocity = 2m.

At t = 4s, s

At t = 5s, s2 = ut + at

S

► velocities u

► heights h

► accelerations - g and g

At highest points, velocities v

Thus, v

⇒ v

and v

⇒ 0 = u

and 0 = u

⇒ u

and u

or u

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