Motion in A Straight Line Test

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Motion in A Straight Line Test - 82

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Weekly Quiz Competition

Question 1
1 / -0

A stone dropped from the top of a towe travels 25m in the last second of its motion g=10m$${ s }^{ -2 }$$ height of the tower is

A
45m

B
90m

C
72m

D
135m

Solution
Question 2
1 / -0

A ball is released from the top of a tower of height h metre. it takes T second to reach the ground, what is the position of the ball in T/3?

A
h/9 m from the ground

B
7h/9 m from the ground

C
8h/9 from the ground

D
17h/18 m from the ground

Solution
Question 3
1 / -0

The acceleration time graph of a particle moving along a straight line is as shown in the figure. After what time the particle acquires $$0$$ velocity.

Note :- The particle initially starts at rest.

A
$$12 \ s$$

B
$$8 \ s$$

C
$$5 \ s$$

D
$$16 \ s$$

Solution
Question 4
1 / -0

The motion of a body falling from rest in a resisting medium is described by the equation $$\dfrac{dv}{dt}=a-bv$$, where a and b are constant. The velocity at any time t is?

A
$$a(1-b^{2t})$$

B
$$\dfrac{a}{b}(1-e^{-bt})$$

C
$$abe^{-t}$$

D
$$ab^2(1-t)$$

Solution
Question 5
1 / -0

A ship moves at 40 km/h due north and suddenly moves towards east through $$9\mathring { 0 } $$ and continuous to move with the same speed. Then the change in its velocity is

A
$$zero$$

B
$$40\sqrt { 2 } $$ km/h north-east

C
$$40\sqrt { 2 } $$ km/h south-east

D
$$None$$

Solution
Question 6
1 / -0

Wind is blowing at a harbor with a speed of 72 km/hr and the flag on the mast of a boat anchored at harbor fluters a long the north -east direction. If the boat starts moving at a speed of $$51 km/hr $$ due north, the direction of the flag is (approximately).

A
Towards east

B
towards west

C
towards north

D
towards south

Solution
Question 7
1 / -0

Directions For Questions

An inquisitive student, determined to test the law of gravity of for himself. walks to the top of a building of $$150$$ floors, with every floor of height $$4\ m$$ having a stopwatch in his hand (the $$1^{st}$$ floor is at a height of $$4\ m$$ from the ground level). From there he jumps off with negligible speed and has start rolling freely. A rocketeer arrive at the scene $$5\ s$$ later and dives off from the top of the building to save the student. The rocketeer leaves the roof with an initial downward speed $$v_0$$. In order to catch the student and to prevent injury to him, the rocketeer should catch the student at a sufficiently great height above ground so that the rocketeer and the student slow down and arrive at the ground with zero velocity. The upward acceleration that accomplishes this is provide by rockteer's jet pack, which he terms on just as he catches the student, before the rocketeer is in free fall. To prevent any discomfort to the atudent, the magnitude of the acceleration of the rocketeer and the student as they move downward together should not exceed $$5\ g$$.

...view full instructions

what would be the approximate retardation to be given by jet pack along for safe landing?

A
$$5g\ ms^{-2}$$

B
$$2g\ ms^{-2}$$

C
$$4g\ ms^{-2}$$

D
$$Cannot\ be\ determined$$
Question 8
1 / -0

A train starts from station with an acceleration $$1 m/s^2$$. A boy who is $$48$$ m behind the train with a constant velocity $$10$$ m/s, the minimum time after which the boy will catch the train is?

A
$$4.8$$ sec

B
$$8$$ sec

C
$$10$$ sec

D
$$12$$ sec

Solution
Question 9
1 / -0

The velocity-time plot for a particle moving on a straight line is shown in the figure, then?

A
The particle has a constant acceleration

B
The particle has never turned around

C
The average speed in the interval $$0$$ to $$10$$ s in the same as the average speed in the interval $$10$$ s to $$20$$ s

D
Both (a) and (c) are correct

Solution

Option A is true because in a velocity-time graph, the slope (which shows the rate of change of velocity) denotes acceleration. In this graph the slope is constant, so acceleration is constant.

Option B is not true because at $$t=10 s$$ the velocity changes its sign from positive to negative, which means it has turned around.

Option C is true because in $$0$$ to $$10 s$$ speed changes from $$10$$ to $$0 \,m/s$$ and in $$10$$ to $$20 s$$ speed changes from $$0$$ to $$10 \,m/s$$. So average speed will be the same.
Question 10
1 / -0

Two particles start from the same point along the same straight line. The first moves with constant velocity '$$v$$' and the second with constant acceleration '$$a$$'. During the time that elapses before the second catch the first, the greatest distance between the particles is:

A
$$ \dfrac{v^{2}}{a} $$

B
$$ \dfrac{v^{2}}{2a} $$

C
$$ \dfrac{2v^{2}}{a} $$

D
$$ \dfrac{v^{2}}{4a} $$

Solution