#### Class nine Science NCERT solutions Chap 8 Motion

**Chapter 8: Motion **

**Intext Questions (PAGE 107)**

**Q.1. What is the nature of the
distance-time graphs for**

** (i) uniform motion **

**(ii) the non-uniform motion of an object?**

**Ans. **(i) The distance-time graph for uniform motion in a straight line not parallel to the time axis.

(ii) The distance-time graph for non-uniform motion is not a straight line. It can be a curve or a zig-zag line not parallel to the time axis.

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

**Ans. **The object is stationary.

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

**Ans. The objects are moving with a uniform motion.**

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

**Ans. **The displacement is measured by the area under a velocity-time graph.

**Intext Questions (PAGE 109)**

**1. A bus starting from rest moves with
a uniform acceleration of 0.1 ms ^{–2} for 2 minutes. Find**

**(a) the speed acquired, (b) the
distance travelled.**

**Ans. **Initial speed (*u*) = 0 (As the
bus is at rest), Final speed (ν) = ? (To be calculated)

Acceleration (*a*) = 0.1 ms^{–2}.
Time (*t*) = 2 min = 120 s

Distance travelled (s) = ? (To be calculated)

(a) Applying, ν = *u *+ *at *= 0 + 0.1 ms^{–2} × 120 s ∴ Final speed (ν) = **12 ms ^{–1}**

Distance
travelled (s) = **720
m**.

**Q.2. A train is travelling at a speed of 90 km h ^{–1}. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s^{–2}. Find how far the train will go before it is brought to rest.**

**Ans. **Initial speed (*u*) = 90 km h^{–1}

Acceleration
(*a*) = – 0.5 ms^{–2}

Distance covered (s) =? (To be calculated)

Applying, ν^{2} – *u*^{2} = 2*as*

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

**Ans. **Initial velocity (*u*) = 0
(zero); Final velocity (ν) = ? (To be calculated)

Time (*t*) = 3 s

Acceleration (*a*) = 2 cms^{–2}

Applying, ν = *u *+ *at *= 0 + 2 cms^{–2}
× 3 s

∴ Final velocity (ν) = **6 cms ^{–1}**

**4. A racing car has a uniform
acceleration of 4 ms–2. What distance will it cover in 10 s, after start?**

**Ans. **Initial velocity (*u*) = 0;
Acceleration (*a*) = 4 ms^{–2}

Time (*t*) = 10 s; Distance covered (s) = ? (To be calculated)

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

**Ans. **Initial velocity of stone (*u*) =
5 ms^{–1}

The final velocity of stone (ν) = 0 (stone comes to rest at some height)

Acceleration in the downward direction = 10 ms^{–2}

∴ Acceleration in the upward direction (*a*)
= –10 ms^{–2}.

Height attained by stone (*s*) = ? (To be
calculated)

Time taken to attain height (*t*) = ? (To
be calculated)

Applying, ν = *u *+ *at *⇒ 0 = 5 ms^{–1} – 10 ms^{–2} × *t*