Motion in A Straight Line Test

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Motion in A Straight Line Test - 45

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Question 1
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A car is moving with speed $$27km/h$$. The driver applied brakes as he approaches a circular turn on the road of radius $$80m$$ and his speed reduces at the constant rate of $$0.50m/s$$ every second. The magnitude of net acceleration is

A
$$20{ms}^{-2}$$

B
$$0.86{ms}^{-2}$$

C
$$100{ms}^{-2}$$

D
None of these

Solution

Speed $$v = 27 \ km/hr = 27\times \dfrac{5}{18} = 7.5\ m/s$$
Radius of circular turn $$r = 80 \ m/s$$
Radial acceleration $$a_r = \dfrac{v^2}{r} = \dfrac{7.5^2}{80} = 0.7 \ m/s^2$$
Tangential acceleration $$a_t = \dfrac{dV}{dt} = 0.50 \ m/s^2$$
Net acceleration $$a_{net} = \sqrt{a_r^2+a_t^2} = \sqrt{0.7^2+0.5^2} = 0.86 \ m/s^2$$
Correct answer is option B.
Question 2
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A body starts from rest and moves with constant acceleration for t s. It travels a distance $$x_1$$ in first half of time and $$x_2$$ in next half of time, then

A
$$x_2 = 3x_1$$

B
$$x_2 = x_1$$

C
$$x_2 = 4x_1$$

D
$$x_2 = 2x_1$$

Solution

$$x_1=0+\dfrac{1}{2}a{\left(\dfrac{t}{2}\right)}^2=\dfrac{at^2}{8}$$ $$[2^{nd}$$ equation of motion]

$$V_B=0+a\left (\dfrac{t}{2}\right)=\dfrac{at}{2}$$ $$[1^{st}$$ equation of motion]

$$x_2=\left (\dfrac{at}{2}\right)\dfrac{t}{2}+\dfrac{1}{2}(a) \left(\dfrac{t}{2}\right)^2$$

$$x_2=\dfrac{at^2}{4}+\dfrac{at^2}{8}=\dfrac{3at^2}{8}$$

$$\Longrightarrow x_2=3x_1$$
Question 3
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A body covers a distance of 4 m in $$3^{rd}$$ second and 12 m in $$5^{th}$$ second. If the motion is uniformly accelerated, how far will it travel in the next 3 seconds?

A
10 m

B
30 m

C
40 m

D
60 m

Solution

We know that distance Travelled by uniformly accelerated body in $$n^{th}s$$ is given by

$$S_n=u+\dfrac{1}{2}a(2n-1),$$ where $$a$$ is the $$acceleration $$ of the body and $$u$$ is the initial velocity

it is given that $$S_n=4s$$ at $$n=3$$
$$\implies 4=u+\dfrac{5a}{2}$$ .............. $$eq(1)$$

$$S_n=12$$ at $$n=5s$$
$$\implies 12=u+\dfrac{9a}{2}$$ ............... $$eq(2)$$

solving $$eq(1)$$ and $$eq(2),$$ we get
$$a=4m/s^2$$ and $$u=-6m/s$$

Now we have to find velocity at $$t=5s$$
we know that $$v=u+at$$
$$\implies v=-6+(4\times 5)=14m/s$$

now distance travelled in next $$3s$$ can be calculated by
$$S=ut+\dfrac{1}{2}at^2$$ $$Note:$$ now $$u=14m/s$$
$$\implies S=14\times 3+\dfrac{1}{2}4(3^2)=60m $$
Question 4
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Two bodies begin a free fall from the same height at a time interval of $$N$$ seconds. If vertical separation between the two bodies is $$1$$ m after n seconds from the start of the first body, then n is equal to

A
$$\sqrt{nN}$$

B
$$\dfrac{1}{gN}$$

C
$$\dfrac{1}{gN}+ \dfrac{N}{2}$$

D
$$\dfrac{1}{gN}- \dfrac{N}{4}$$

Solution

Let $$x_1$$ be the distance travelled by the first body.
Let $$x_2$$ be the distance travelled by the second body.
Using $$s=ut+\dfrac{1}{2}gt^2$$,
$$\Rightarrow -x_1= 0\times n -\dfrac{1}{2} gn^2$$ $$\Rightarrow x_1= \dfrac{1}{2}gn^2$$,
$$\Rightarrow -x_2= 0\times (n-N) -\dfrac{1}{2} g(n-N)^2$$ $$\Rightarrow x_2= \dfrac{1}{2}g(n-N)^2$$,
Given $$x_1-x_2=1=\dfrac{1}{2}gn^2-\dfrac{1}{2} g(n-N)^2$$ $$\Rightarrow n= \dfrac{1}{gN}+\dfrac{N}{2}$$
Question 5
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Which of the following statements is incorrect ?

A
In one dimension motion, the velocity and the acceleration of an object are always along the same line.

B
In two or three dimensions, the angle between velocity and acceleration vectors may have any value between 0$$^{\circ}$$and 180$$^{\circ}$$

C
The kinematic equations for uniform acceleration can be applied in case of uniform circular motion.

D
The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant.

Solution

The kinematic equations for uniform acceleration do not apply in case of uniform circular motion because in this case the magnitude of acceleration is constant but its direction is changing.
Question 6
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A body $$A$$ starts from rest with an acceleration $$a_1$$. After $$2$$ seconds, another body $$B$$ starts from rest with an acceleration $$a_2$$. If they travel equal distances in the $$5^{th}$$ second, after the start of $$A$$, then the ratio $$a_1$$ : $$a_2$$ is equal to

A
$$5 : 9$$

B
$$5 : 7$$

C
$$9 : 5$$

D
$$9 : 7$$

Solution

Initial speed $$u = 0$$
Distance covered in nth second after starting from rest   $$S = \dfrac{1}{2}(2n-1)$$
For A :   $$n = 5$$
So,   $$S_A = \dfrac{a_1}{2}(2\times 5-1) = \dfrac{9}{2}a_1$$
For B :   $$n = 3$$
So,   $$S_B = \dfrac{a_2}{2}(2\times 3-1) = \dfrac{5}{2}a_2$$
But   $$S_A=S_B$$
Or   $$\dfrac{9}{2}a_1 = \dfrac{5}{2}a_2$$
$$\implies \ a_1:a_2 = 5:9$$
Question 7
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An aircraft is flying at a height of $$3400 \,m$$ above the ground. If the angle subtended at a ground observation point by the aircraft positions $$10 \,s$$ apart is $$30^o$$, then the speed of the aircraft is:

A
10.8 $$ms^{-1}$$

B
1963 $$ms^{-1}$$

C
108 $$ms^{-1}$$

D
196.3 $$ms^{-1}$$

Solution

O is the observation point at the ground. A and B are the positions of aircraft for which $$\angle$$AOB = 30$$^{\circ}$$.Time taken by aircraft from A to B is 10 s.
In $$\Delta$$ AOB,
$$tan 30^{\circ}=\dfrac{AB}{3400}; \ \ \ AB= 3400 \ tan 30^{\circ}=\dfrac{3400}{\sqrt{3}}$$m
$$\therefore$$ Speed of aircraft, $$v = \dfrac{AB}{10}=\dfrac{3400}{10\sqrt{3}} = 196.3 ms^{-1}$$
Question 8
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A lift is coming from $$8^{th}$$ floor and is just about to reach $$4^{th}$$ floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?

A
x < 0, v < 0, a > 0

B
x > 0, v < 0, a < 0

C
x > 0, v < 0, a > 0

D
x < 0, v < 0, a = 0

Solution

$$v$$ is $$-ve$$ $$(v<0)$$
$$x(displacement)$$ is $$-ve$$ $$(x<0)$$
Lift is about to reach $$4^{th}$$ floor, here $$v=0\Longrightarrow v \quad is \quad \downarrow$$
$$\Longrightarrow a$$ is in opposite direction of $$v.$$
$$\Longrightarrow a=+ve.$$
Question 9
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On a long horizontally moving belt, a child runs to and fro with a speed 9 km $$h^{-1}$$ (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km $$h^{-1}$$. For an observer on a stationary platform, the speed of the child running in the direction of motion of the belt is

A
4 km $$h{-1}$$

B
5 km $$h{-1}$$

C
9 km $$h{-1}$$

D
13 km $$h{-1}$$

Solution

$${ V }_{ 1 }\rightarrow $$ Velocity of child wrt ground
$${ V }_{ 2 }\rightarrow $$ Velocity of belt wrt ground
$${ V }_{ child/belt }=9 \\{ V }_{ belt/g }=4$$

$${ V }_{ child/belt }={ V }_{ child/go }-{ V }_{ belt/g }$$
$$9={ V }_{ 1 }-4$$
$$\Rightarrow \boxed { { V }_{ 1 }=13km/hr } $$ [both are in same direction].
Question 10
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A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plots shown in figure, indicate the one that represents the velocity ($$v$$) of the pebble as a function of time ($$t$$).

A

B

C

D

Solution

When the pebble is dropped from the top of cylinder filled with viscous oil, pebble falls under gravity with constant acceleration, but as it is dropped it enters in oil and dragging force $$F = 6\eta \nu r v$$ due to viscosity of oil so acceleration decreases from g to zero i.e. velocity increases, but acceleration decreases, when acceleration decreased to zero, velocity becomes constant (terminal velocity). These conditions are verified in option ($$c$$).