Motion in A Straight Line Test

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Motion in A Straight Line Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

A body thrown vertically up from the ground passes the height $$10.2m$$ twice at an interval of $$10 sec$$. Its initial velocity was (in m/s) :

A
$$52$$

B
$$26$$

C
$$51$$

D
$$20$$

Solution

Here the time 5 sec is for the body to rise from 10.2 m elevation to the maximum height and another 5 sec is for fall back to 10.2 m elevation.
Since at highest point the velocity of the body is zero, so the velocity at 10.2 m height is $$v_h=gt=10(5)=50 m/s$$ (using $$v=u+at$$ here $$v=0, u=v_h, a=-g$$)
Using conservation of energy, $$mgh=\dfrac{1}{2}m[v_0^2-v_h^2]$$
or $$v_0^2=2gh+50^2=2(10)(10.2)+50^2$$
or $$v_0=52 m/s$$
Question 2
1 / -0

A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is _____
(Assume, $$g = 10 m{s}^{-2}$$)

A
$$5 m$$

B
$$10 m$$

C
$$7.5 m$$

D
$$20 m$$

Solution

The person throws balls into air vertically upward in regular intervals of time of one second.And next ball is thrown when the velocity of the ball thrown earlier becomes zero.
So time to reach the highest point is 1 sec. If initial velocity is u then at highest point,
$$0=u-gt$$
$$t=1s$$ $$g=10m/s^2$$
$$u=10m/s$$
Height (H) to which ball rises ,
0=$$u^2$$-2gH
H=$$\dfrac{u^2}{2g}$$=5m
Question 3
1 / -0

Find the magnitude of constant acceleration (in $$m/s^2$$) needed to allow a car to accelerate in a straight line from a speed of zero to a speed of $$30 m/s$$ in $$5 s$$ :

A
$$7$$

B
$$6$$

C
$$5$$

D
$$4$$

Solution

Here acceleration , $$a=\dfrac{dv}{dt}=\dfrac{30-0}{5}=6 m/s^2$$
Question 4
1 / -0

A motor scooter travels east at a speed of $$13 m/s$$. The driver then reverses direction and heads west at $$17 m/s$$. What was the change in velocity of the scooter? (in m/s)

A
$$4$$

B
$$30$$

C
$$32$$

D
$$21$$

Solution

Consider the direction of west to be positive direction.
Final velocity $$v = 17 \ m/s$$
Initial velocity $$u = -13 \ m/s$$
Thus change in velocity $$\Delta v = v-u = 17-(-13) = 30 \ m/s$$
Question 5
1 / -0

An airplane must reach a take of speed of $$80 m/s$$ in a $$1000 m$$ long runway. What minimum constant acceleration is required? (in $$m/s^2$$)

A
$$3.2$$

B
$$2.2$$

C
$$1$$

D
$$4$$

Solution

Here, initial velocity is $$u=0 m/s$$ (as airplane starts from rest)
Final velocity is $$v=80 m/s$$ and distance traveled, $$S=1000 m$$.
Let $$a$$ be the required acceleration.
Using formula $$v^2-u^2=2aS$$,
$$(80)^2-0^2=2a(1000)$$
So, we get $$a=3.2 m/s^2$$
Question 6
1 / -0

A race car accelerates uniformly from $$18.5 m/s$$ to $$46.1 m/s$$ in $$2.47$$ seconds. Determine the acceleration of the car. (in $$m/s^2$$)

A
$$11.2 m/s^2$$

B
$$11 m/s^2$$

C
$$1.2 m/s^2$$

D
$$51.2 m/s^2$$

Solution

Initial velocity of the car   $$u = 18.5 \ m/s$$
Final velocity of the car $$v = 46.1 \ m/s$$
Time taken   $$t = 2.47 \ s$$
Acceleration of the car   $$a = \dfrac{v-u}{t}$$
$$\implies \ a = \dfrac{46.1-18.5}{2.47} = 11.2 \ m/s^2$$
Question 7
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Directions For Questions

A rock is dropped from rest into a well.The sound of the splash is heard $$4 s$$ after the rock is released from rest.(the speed of sound in air at ambient temperature is $$336m/s$$).

...view full instructions

How far below to top of the well is the surface of the water?

A
$$72 m$$

B
$$71 m$$

C
$$70.4 m$$

D
$$70 m$$

Solution

Total time taken= Time taken by stone to hit tge water surface($$t_1$$)+Time taken by sound wave to reach back to the ears($$t_2$$)
$$\Rightarrow 4=t_1+t_2$$
Let the depth of water surface from top of well be d
$$\Rightarrow t_1=\sqrt{\dfrac{2d}{g}}$$
Also, $$t_2=\dfrac{d}{336}$$
$$\Rightarrow 4=\sqrt{\dfrac{2d}{g}}+\dfrac{d}{336}$$
$$\Rightarrow d=70.4$$

Hence,correct answer is option $$C$$
Question 8
1 / -0

Luke Autbeloe drops a pile of roof shingles from the top of a roof located $$8.52$$ meters above the ground. Determine the time required for the shingles to reach the ground.

A
$$1 s$$

B
$$1.32 s$$

C
$$1.3 s$$

D
$$1.2 s$$

Solution

Here, the acceleration of shingles is $$a=g=9.8 m/s^2$$,
Initial velocity of it is $$u=0 m/s$$ and distance traveled $$S=8.52 m$$
Let $$t$$ be the required time for the shingles to reach ground.
Using formula $$S=ut+at^2/2$$
we get, $$8.52=0(t)+0.5(9.8)t^2$$
$$\implies$$ $$t=1.32 s$$
Question 9
1 / -0

A marble starts falling from rest on a smooth inclined plane forming an angle $$\alpha$$ with horizontal. After covering distance $$h$$ the ball rebound off the plane. The distance from the impact point where the ball rebounds for second time is

A
$$8h sin{\alpha}$$

B
$$8h cos{\alpha}$$

C
$$h sin{\alpha}$$

D
$$h cos{\alpha}$$

Solution

The balls fall from $$A$$ on to incline and bounces off to $$B$$
$$B{ B }^{ ' }$$ is the normal to the incline at $$B$$.
Since it is smooth plane and collision l bouncing is elastic, the ball bounces at the same angle with the normal $$B{ B }^{ ' }$$
The velocity remains same
Hence the angle of projection of the ball w.r.t. horizontal$$=90°2\alpha $$
Velocity of the ball at $$B=\nu =\sqrt { 2gh } $$
Let $$B=(0,0)$$
The equation of the trajectory of the ball (parabolic path)
$$y=x\tan { \left( 90°-2\alpha \right) } -\cfrac { g{ x }^{ 2 } }{ 2{ \nu }^{ 2 } } \sec ^{ 2 }{ (90°-2\alpha ) } $$
Equations of the inclined plane: $$y=-x\tan { \alpha } $$
Ball hits the plane at
$$-x\tan { \alpha } =x\cot { 2\alpha } -\cfrac { g{ x }^{ 2 } }{ 4gh } \csc ^{ 2 }{ \left( 2\alpha \right) } $$
$$\therefore x=0$$ or $$x=\left( \cot { 2\alpha } +\tan { \alpha } \right) 4h\sin ^{ 2 }{ 2\alpha } $$
or $$c=\cfrac { 4h\left( 1-\tan ^{ 2 }{ \alpha } +2\tan ^{ 2 }{ \alpha } \right) \sin ^{ 2 }{ 2\alpha } }{ 2\tan { \alpha } } $$
$$x=4h\sin { 2\alpha } $$
$$y=-4h\sin { 2\alpha } \cdot \tan { \alpha } =-8h\sin ^{ 2 }{ 2\alpha } $$
Distance$$=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =8h\sin { \alpha } $$
The distance on inclined plane from $$B$$ (point of first bounce to point of second bounce)$$=\cfrac { x }{ \cos { \alpha } } =8h\sin { \alpha } $$
Question 10
1 / -0

A flowerpot falls from a windowsill $$25.0 m$$ above the sidewalk. How fast is the flowerpot moving when it strikes the ground? (in m/s)

A
$$22$$

B
$$23$$

C
$$26$$

D
$$25$$

Solution

Initial speed of flowerpot $$u=0$$
Height $$h = 25.0 \ m$$
Applying $$3_{rd}$$ equation of motion,
$$v^{2}=u^{2}+2 gh$$
$$v^{2}=0^{2}+ 2 \times 9.8 \times25$$
$$v=22 \ m/s$$

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 40

Result

Motion in A Straight Line Test - 40

Score

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Rank

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SHARING IS CARING

Question 1

$$52$$

$$26$$

$$51$$

$$20$$

Solution

Question 2

$$5 m$$

$$10 m$$

$$7.5 m$$

$$20 m$$

Solution

Question 3

$$7$$

$$6$$

$$5$$

$$4$$

Solution

Question 4

$$4$$

$$30$$

$$32$$

$$21$$

Solution

Question 5

$$3.2$$

$$2.2$$

$$1$$

$$4$$

Solution

Question 6

$$11.2 m/s^2$$

$$11 m/s^2$$

$$1.2 m/s^2$$

$$51.2 m/s^2$$

Solution

Question 7

$$72 m$$

$$71 m$$

$$70.4 m$$

$$70 m$$

Solution

Question 8

$$1 s$$

$$1.32 s$$

$$1.3 s$$

$$1.2 s$$

Solution

Question 9

$$8h sin{\alpha}$$

$$8h cos{\alpha}$$

$$h sin{\alpha}$$

$$h cos{\alpha}$$

Solution

Question 10

$$22$$

$$23$$

$$26$$

$$25$$

Solution

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