Motion in A Straight Line Test

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Motion in A Straight Line Test - 73

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following equations represents the motion of a body moving with constant finite acceleration? in these equations, $$y$$ denote the displacement in time $$t$$ and $$p, q$$ and $$r$$ are arbitary constants:

A
$$y = (p + qt)^2(r + pt)$$

B
$$y = p + tqr$$

C
$$y = (p + t)(q + t)(r + 1)$$

D
$$y = (p + qt)r$$

Solution
Question 2
1 / -0

A body is projected vertically upwards with speed $$40 m/s$$ . The distance traveled by the body in the last second of upward journey is ( take g= $$9.8 m/s^2$$ and neglect effect of air resistance)

A
4.9 m

B
9.8 m

C
12.4 m

D
19.6 m

Solution

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Question 3
1 / -0

A trolley is moving horizontally with a constant velocity of v m/s w.r.t. earth. A man starts running from one end of the trolley with a velocity $$1.5v m/s$$ w.r.t. to trolley. After reaching the opposite end, the man return back to starting end and continues running with a velocity of 1.5 v m/s w.r.t. the trolley in the backward direction. If the length of the trolley is L then the displacement of the man with respect to earth during the process will be :

A
$$2.5 L$$

B
$$1.5 L$$

C
$$\dfrac { 5L }{ 3 } $$

D
$$\dfrac { 4L }{ 3 } $$

Solution
Question 4
1 / -0

A body is thrown vertically upwards and takes 5 seconds to reach maximum height. the distance travelled by the body will be same in

A
1st and 10th second

B
2nd and 8th second

C
4th and 6th second

D
both (b) and (c)

Solution

Hint:
The The distance travelled by the particle is always same for just before and just after second of reaching the maximum height.

Step 1: Explanation,
$$\bullet$$The body reaches the maximum height at $$5 sec$$.
The just before and just after seconds are $$5^{th}$$ and $$6^{th} sec $$ respectively are equal. So option C is incorrect.

$$\bullet$$The total time of flight is 10 sec .so distance between 1st means (0-1sec) and 10th means (9-10sec) are equal.

$$\bullet$$ The distance between 2nd means (1 sec-2 sec) and 9th means (8 sec -9sec) are equal.

Therefore option A is the correct answer.
Question 5
1 / -0

A taxi leaves the station $$X$$ for station $$Y$$ every $$10$$ minutes. Simultaneously, a taxi also leaves the station $$Y$$ for station $$X$$ every $$10$$ minutes. The taxies move at the same constant speed and go from $$X$$ to $$Y$$ or vice versa in $$2$$ hours. How many taxies coming from the other side will meet each taxi in route from $$Y$$ to $$X$$?

A
$$10$$

B
$$11$$

C
$$12$$

D
$$23$$
Question 6
1 / -0

Two cars are moving in the same directions with the same speed of $$30$$ km/hr. They are separated by $$5$$ km. What is the speed of the car moving in the opposite direction if it meets the second cars at an interval of $$4$$ minute?

A
$$45$$ km/hr

B
$$60$$ km/hr

C
$$105$$ km/hr

D
None of these

Solution
Question 7
1 / -0

A particle moves in a straight line with a constant acceleration. it changes its velocity from $$10\ ms^{-1}$$ to $$20\ ms^{-1}$$ while passing through a distance $$135\ m$$ in $$t$$ seconds. The value of $$t$$ is

A
$$9$$

B
$$10$$

C
$$1.8$$

D
$$12$$

Solution

Given,
$$u=10m/s$$
$$v=20m/s$$
$$S=135m$$
$$t=?$$
From 3rd equation of motion,
$$2as=v^2-u^2$$
$$a=\dfrac{v^2-u^2}{2s}$$
$$a=\dfrac{20\times 20-10\times 10}{2\times 135}$$
$$a=1.11m/s^2$$
From 1st equation of motion,
$$v=u+at$$
$$20=10+1.11t$$
$$10=1.11t$$
$$t=\dfrac{10}{1.11}$$
$$t=9sec$$
The correct option is A.
Question 8
1 / -0

A particle starts from rest with constant acceleration for 20 s. If it travels a distance $$y_1$$ in the first 10 s and a distance $$y_2$$ in the next 10 s then

A
$$y_2=2y_1$$

B
$$y_2=3y_1$$

C
$$y_2=4y_1$$

D
$$y_2=5y_1$$

Solution
Question 9
1 / -0

A body is allowed to slide from the top along a smooth inclined plane of length 5m at an angle of inclination $$30^o$$. If $$g = 10ms^{-2}$$, time taken by the body to reach the bottom of the plane is

A
$$\dfrac{\sqrt{3}}{2}s$$

B
$$1.41s$$

C
$$\dfrac{1}{\sqrt{2}}s$$

D
$$2s$$

Solution
Question 10
1 / -0

A parachutist after bailing out falls for $$10s$$ without friction. When the parachute opens he descends with an acceleration of $$2\ m/s^{2}$$ against his direction and reached the ground with $4\ m/s$. From what height he has dropped himself? $$(g=10m/s^{2})$$

A
$$500m$$

B
$$2496m$$

C
$$2996m$$

D
$$4296m$$

Solution