Chapter 07 Motion Questions-05
Questions
1. A bus starting from rest moves with a uniform acceleration of $0.1 m s^{-2}$ for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
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Answer
Initial speed of the bus, $u=0$
Acceleration, $a=0.1 m / s^{2}$
Time taken, $t=2$ minutes $=120 s$
(a) $v=u+a t$
$v=0+0 \times 1 \times 120$
$v=12 ms^{-1}$
(b) According to the third equation of motion:
$v^{2}-u^{2}=2 a s$
Where, $s$ is the distance covered by the bus
$(12)^{2}-(0)^{2}=2(0.1) s$
$s=720 m$
Speed acquired by the bus is $12 m / s$.
Distance travelled by the bus is $720 m$.
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.
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Answer
Initial speed of the train, $u=90 km / h=25 m / s$
Final speed of the train, $v=0$ (finally the train comes to rest)
Acceleration $=-0.5 m s^{-2}$
According to third equation of motion:
$v^{2}=u^{2}+2$ as
$(0)^{2}=(25)^{2}+2(-0.5) s$
Where, $s$ is the distance covered by the train
$ s=\frac{25^{2}}{2(0.5)}=625 m $
The train will cover a distance of $625 m$ before it comes to rest.
3. 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?
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Answer
Initial Velocity of trolley, $u=0 cms^{-1}$
Acceleration, $a=2 cm s^{-2}$
Time, $t=3 s$
We know that final velocity, $v=u+a t=0+2 \times 3 cms^{-1}$
Therefore, The velocity of train after 3 seconds $=6 cms^{-1}$
4. A racing car has a uniform acceleration of $4 m s^{-2}$. What distance will it cover in $10 s$ after start?
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Answer
Initial Velocity of the car, $u=0 ms^{-1}$
Acceleration, $a=4 m s^{-2}$
Time, $t=10 s$
We know Distance, $s=u t+(1 / 2) a t^{2}$
Therefore, Distance covered by car in 10 second $=0 \times 10+(1 / 2) \times 4 \times$ 102
$=0+(1 / 2) \times 4 \times 10 \times 10 m$
$=(1 / 2) \times 400 m$
$=200 m$
5. 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 $m s^{-2}$ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
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Answer
Given Initial velocity of stone, $u=5 m s^{-1}$
Downward of negative Acceleration, $a=10 m s^{-2}$
We know that $2 a s=v^{2}-u^{2}$
Therefore, Height attained by the stone, $s=\frac{0^{2}}{5^{2}} \times(-10) m$
$=\frac{-25}{-20} m$
$=1.25 m$
Also we know that final velocity, $v=u+at$
or, Time, $t=\frac{v-u}{a}$
Therefore, Time, $t$ taken by stone to attain the height, $s=\frac{0-5}{-10 s}$ $=0.5 s$
BETA
