Motion in A Straight Line Test

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Motion in A Straight Line Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

A plane has a takeoff speed of $$88.3 m/s$$ and requires $$1365 m$$ to reach that speed. Determine the time required to reach this speed.

A
$$20. 8 s$$

B
$$30. s$$

C
$$30. 8 s$$

D
$$40. 8 s$$

Solution

Given, initial velocity is $$u=0 m/s$$
Final velocity is $$v=88.3 m/s$$ and traveled distance, $$d=1365 m$$
Let $$a$$ be the acceleration of the plane.
Using $$v^2-u^2=2ad$$
$$\therefore$$ $$(88.3)^2-0^2=2a(1365)$$
$$\implies$$ $$a=2.86 m/s^2$$
If $$ t$$ be the required time.
Using $$v=u+at$$,
$$88.3=0+2.86t$$
$$\implies$$ $$t=88.3/2.86=30.8 s$$
Question 2
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Directions For Questions

A stone is thrown vertically upward was noted having a velocity of $$15 m/s$$ after covering $$2/3$$ of its maximum height.

...view full instructions

What is the total height reached by the stone? (in m)

A
$$36$$

B
$$35$$

C
$$34.44$$

D
$$39$$

Solution

Maximum height in verticle motion is given by $$H=\frac{u^2}{2g}$$
Also by using $$v^2-u^2=2aS$$
Given v=15, S=$$\dfrac{2H}{3}$$, using $$a=g=-9.81$$
We get $$u^2=225+13.08H$$
Solving both equations eleminating $$u^2$$
We get
$$2gH=225+13.08H$$
$$H=34.403$$
So closest option is optionC.
Question 3
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It was once recorded that a Jaguar left skid marks that were $$290 m$$ in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of $$-3.90 m/s^2$$, determine the speed of the Jaguar before it began to skid.

A
$$47.6 m/s$$

B
$$38.2 m/s$$

C
$$54.6 m/s$$

D
$$57.6 m/s$$

Solution

Final velocity   $$v = 0$$
Acceleration of Jaguar   $$a = -3.9 \ m/s^2$$
Distance covered $$s = 290 \ m$$
Using   $$v^2 - u^2 = 2aS$$
$$\therefore \ 0-u^2 = 2(-3.9)(290)$$
$$\implies \ u = \sqrt{2(3.9)(290)} = 47.6 \ m/s$$
Question 4
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An aeroplane is flying horizontally with a velocity of $$600 km/h$$ and at a height of $$1960 m$$. When it is vertically at a point $$A$$ on the ground a bomb is released from it. The bomb strikes the ground at point $$B$$. The distance $$AB$$ is:

A
$$1200 m$$

B
$$0.33 km$$

C
$$3.33 km$$

D
$$33 km$$

Solution

The velocity with which the bomb is released is $$600km/h(horizontal)$$.
Vertical velocity of the bomb is $$zero$$.
So.distance $$AB=600\dfrac{5}{18}*t$$ $$t$$ is the time taken by the bomb to reach the ground.
$$1960=g\dfrac{{t}^{2}}{2}$$
which gives $$AB$$ equal to $$3.3.km$$.
Question 5
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The K.E. (K) of a particle moving along a circle of radius $$R$$ depends on the distance covered $$s$$ as $$K = a s^2$$. The force acting on particle is

A
$$\dfrac{2as^{2}}{R}$$

B

C

D
none of these.

Solution

Given $$K.E=\cfrac { 1 }{ 2 } m{ v }^{ 2 }=a{ s }^{ 2 }$$
$$\therefore m{ v }^{ 2 }=2a{ s }^{ 2 }$$
differentiating w.r.t.
$$2mv\cfrac { dv }{ dt } =4as\cfrac { ds }{ dt } =4asv$$
$$mat=2as$$
Here $${ F }_{ t }=2as$$ ($${ F }_{ t }=$$tangential force)-$$1$$
Centrifugal force$$={ F }_{ c }={ m }_{ ac }=\cfrac { m{ v }^{ 2 } }{ R } $$
$$=\cfrac { 2ma{ s }^{ 2 } }{ mR } =\cfrac { 2a{ s }^{ 2 } }{ R } \rightarrow 2$$
Net force on particle is
$$=\sqrt { { F }_{ t }^{ 2 }+{ F }_{ c }^{ 2 } } =\sqrt { { (2as) }^{ 2 }+{ \left( \cfrac { 2a{ s }^{ 2 } }{ R } \right) }^{ 2 } } $$
$$=2as\sqrt { 1+\cfrac { { s }^{ 2 } }{ { R }^{ 2 } } } $$
$$=2as{ \left( 1+\cfrac { { s }^{ 2 } }{ { R }^{ 2 } } \right) }^{ { 1 }/{ 2 } }$$
Question 6
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A car accelerates from rest at constant rate of $$2 m/s^2$$ for some time. Then its retards at a constant rate of $$4 m/s^2$$ and comes to rest. if it remains in motion for $$3$$ seconds, then the maximum speed attained by the car is:-

A
$$2 m/s$$

B
$$3 m/s$$

C
$$4 m/s$$

D
$$6 m/s$$

Solution

Let $$v, t_1,t_2$$ be the maximum velocity attained, time for which car accelerated and the time for which car retarded respectively
$$\Rightarrow t_1=(v-u)/a=(v-0)/2=v/2$$
$$\Rightarrow t_2=(v-u)/a=(0-v)/(-4)=v/4$$

Now it is given that,
$$\Rightarrow t_1+t_2=3$$
$$\Rightarrow v/2+v/4=3$$
$$\Rightarrow v=4m/s$$

Hence correct answer is option $$C $$
Question 7
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A particle is projected vertically upwards and it attains maximum height $$H$$. If the ratio of times to attain height $$h(h < H)$$ is $$1/3$$, then $$h$$ equals

A
2/3 H

B
3H

C
5/9. H

D
3/. H

Solution

Total time taken to reach height H$$=t=\sqrt{\dfrac{2H}{g}}$$
Initial velocity$$=u=\sqrt{2gH}$$
Time to reach height h$$=t_1=t/3=\dfrac{1}{3} \sqrt{\dfrac{2H}{g}}$$

Now by second equation of motion
$$\Rightarrow h=ut_1+0.5\times(-g)(t_1^2)$$

$$\Rightarrow h=\sqrt{2gH}\times $$$$\dfrac{1}{3}\times \sqrt{\dfrac{2H}{g}}$$$$-0.5\times g \times {(\dfrac{1}{3} \sqrt{\dfrac{2H}{g}})}^2$$

$$\Rightarrow h=\dfrac{2H}{3}-\dfrac{H}{9}=\dfrac{5H}{9}$$

$$\Rightarrow \dfrac{h}{H}=\dfrac{5}{9}$$

Hence none of the options are correct.
Question 8
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Directions For Questions

A ball is thrown vertically upward at $$20 m/s$$ ignoring air resistance and taking g as $$9.8 m/s^2$$ . Calculate:

...view full instructions

Maximum height. (in m)

A
$$20$$

B
$$40$$

C
$$60$$

D
$$80$$

Solution

Let the maximum height reached be $$H$$.
At the maximum height, final velocity of ball is zero i.e. $$v = 0$$
Initial velocity $$u = +20 \ m/s$$ (Considering upward direction to be positive)
Acceleration due to gravity $$g = -9.8 \ m/s^2$$
Using $$v^2 - u^2 =2gH$$
Or $$0-20^2 = 2(-9.8)H$$
$$\implies\ H \approx 20 \ m$$
Question 9
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From the top of a tower, a stone is thrown up. It reaches the ground in $$5$$ $$s$$. A second stone is thrown down with the same speed and reaches the ground in $$1$$ $$s$$. A third stone is released from rest and reaches the ground in

A
3 $$s$$

B
2 $$s$$

C
$$\sqrt 5$$ $$s$$

D
2.5 $$s$$

Solution

let height of tower is $$H$$, acceleration due to gravity is $$a$$,
case 1) initial velocity is $$-u$$ (assuming downward direction as positive.
then $$H=-ut+\frac{1}{2}at^2 =>H=-5u+\frac{25}{2}a$$
for case 2) initial velocity is $$u$$
then $$H=ut+\frac{1}{2}at^2=>H=u+\frac{1}{2}a$$
multiplying 2nd with 5 and adding both equation we get
$$6H=15a=>H=2.5a$$
now for 3rd case $$u=0$$, we get $$2.5a=\frac{1}{2}at^2=>t=\sqrt5 sec.$$
Question 10
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Two particles P and Q start from rest and move for equal time on a straight line. Particle P has an acceleration of$$X m/s^2$$ for the first half of the total time and $$2x m/s^2$$ for the second half. Particle Q has an acceleration of $$2X m/s^2$$ for the first half of the total time and $$X m/s^2$$ for the second half. Which particle has covered larger distance?

A
Both have covered the same distance

B
P has covered the larger distance

C
Q has covered the larger distance

D
data insufficient

Solution

assuming total time is $$2t$$
for $$P$$, at half time
$$v=X.t, S_1=\frac{1}{2}Xt^2$$
at full time $$S_2=Xt.t+\frac{1}{2}2Xt^2=2Xt^2$$
total distance covered by $$P$$ is $$ S_1+S_2=2.5Xt^2$$
for particle $$Q$$, at half time
$$v=2Xt,S_1=\frac{1}{2}2X t^2=Xt^2$$
at full time $$S_2=2Xt.t+\frac{1}{2}Xt^2=2.5Xt^2$$
total distance covered by $$Q$$ is $$ S_1+S_2=3.5Xt^2$$
so distance covered by $$Q $$ is more then $$P$$.
correct answer is C.

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Motion in A Straight Line Test - 41

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Motion in A Straight Line Test - 41

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Question 1

$$20. 8 s$$

$$30. s$$

$$30. 8 s$$

$$40. 8 s$$

Solution

Question 2

$$36$$

$$35$$

$$34.44$$

$$39$$

Solution

Question 3

$$47.6 m/s$$

$$38.2 m/s$$

$$54.6 m/s$$

$$57.6 m/s$$

Solution

Question 4

$$1200 m$$

$$0.33 km$$

$$3.33 km$$

$$33 km$$

Solution

Question 5

$$\dfrac{2as^{2}}{R}$$

none of these.

Solution

Question 6

$$2 m/s$$

$$3 m/s$$

$$4 m/s$$

$$6 m/s$$

Solution

Question 7

2/3 H

3H

5/9. H

3/. H

Solution

Question 8

$$20$$

$$40$$

$$60$$

$$80$$

Solution

Question 9

3 $$s$$

2 $$s$$

$$\sqrt 5$$ $$s$$

2.5 $$s$$

Solution

Question 10

Both have covered the same distance

P has covered the larger distance

Q has covered the larger distance

data insufficient

Solution

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