Motion in A Straight Line Test

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Motion in A Straight Line Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

A stone is thrown vertically upward with a velocity of 34.3 m/s from the top of a cliff 147 m high . after it reaches the foot of the cliff, the mine elapsed will be :

A
3 sec

B
13 sec

C
7 sec

D
10 sec

Solution

Correct option is $$(D)$$

As we know;
$$ s=u t+\frac{1}{2} a t^{2} $$
$$ \begin{array}{l}-147=(34.3) t+\frac{1}{2}(-9.8) t^{2} \\ -147=(34.3) t-(4.9) t^{2} \\ \text { Hence, }(4.9) t^{2}-(34.3) t-147=0 \\ \text { By Solving this we get; } \\ t=10\end{array} $$
Question 2
1 / -0

Water drugs full from a tap on the floor 5 m below at regular intervals of time, The first drop beings to fall. The height at which the third drop will be from ground (at the instant when the first drop strikes the ground) will be $$\left( g=10{ ms }^{ -2 } \right) $$

A
1.25 m

B
2.15 m

C
2.75 m

D
3.75 m

Solution
Question 3
1 / -0

When a body is accelerated: (i) its velocity always changes (ii) its speed always changes (iii) its direction always changes (iv) its speed may or may not change Which of the following is correct?

A
(i) and (ii)

B
(i) and (iv)

C
(i) only

D
(ii) and (iii)

Solution
Question 4
1 / -0

A stone falls from a balloon that is descending at a uniform rate of 12m/s. the displacement of the stone from the point of release 10 sec is

A
490 m

B
510 m

C
610 m

D
725 m

Solution
Question 5
1 / -0

An object falls freely from rest in a vacuum. The graph shows the variation with time t of the
velocity $$v$$ of the object.
Which graph, using the same scales, represents the object falling in air?

A

B

C

D

Solution

In free fall in air, the only force acting on object is gravity.
$$\dfrac{dv}{dt}= g$$
Since $$g$$ is constant.
$$\implies \dfrac{dv}{dt}= $$constant
This implies that slop of $$v-t$$ graph is constant.
So, option A is correct.
Question 6
1 / -0

In the two dimensional motions:

A
$$x-t$$ graph gives actual path of the particle

B
$$y-t$$ graph gives actual path of the particle

C
$$\sqrt{x^2+y^2}$$ versus t graph gives the actual path of the particle

D
$$y-x$$ graph gives actual path of particle

Solution

Motion in two dimensions:
Motion in a plane is described as two-dimensional motion.
Example: Projectile and circular motion are examples of two-dimensional motion.
Hence, Path followed by the particle in 2D two variables (x-y).
option D
Question 7
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An object has an initial velocity $$u$$ and an acceleration $$a$$. The object moves in a straight line
through a displacement s and has final velocity $$v$$.
The above quantities are related by the equation shown.
$$v^{2} = u^{2} + 2as$$
Which condition must be satisfied in order for this equation to apply to the motion of the object?

A
The direction of $$a$$ is constant and the direction of $$a$$ is the same as the direction of $$s$$

B
The direction of $$a$$ is constant and the direction of $$a$$ is the same as the direction of $$u$$

C
The magnitude of $$a$$ is constant and the direction of $$a$$ is constant

D
The magnitude of $$a$$ is constant and the direction of $$a$$ is the same as the direction of $$v$$

Solution

$$v^{2} = u^{2} + 2as$$
The above equation of motion is valid only for constant acceleration.
So, for constant acceleration , the magnitude and direction of acceleration must be constant.
Question 8
1 / -0

The velocity-time graph of a body is shown in Fig.4.116. The displacement of the body in 8 s is

A
9 $$m$$

B
12 $$m$$

C
10 $$m$$

D
28 $$m$$

Solution

Displacement = area under graph
$$Area=\quad 2\times 2+\dfrac { 1 }{ 2 } \left( 2+6 \right) \times 1+\dfrac { 1 }{ 2 } \times 1\times 6-\dfrac { 1 }{ 2 } +1\times 6-1\times 6+2\times 4$$
$$=4+4+3-3-6+8=10m$$
$$Area=10m=displacement \ in \ 8sec.$$
Question 9
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Two particles A and B move from rest along a straight line with constant accelerations f and h respectively. If A takes m seconds more than B and describes n units more than that of B acquiring the same speed, then?

A
$$(f+h)m^2=fhn$$

B
$$(f-fh)m^2=fhn$$

C
$$(h-f)n=\dfrac{1}{2}fhm^2$$

D
$$\dfrac{1}{2}(f+h)n=fhm^2$$

Solution

$$S+n=\dfrac{1}{2}f(t+m)^2$$ and $$S=\dfrac{1}{2}ht^2, V=ht$$
$$\therefore \dfrac{1}{2}ht^2+n=\dfrac{1}{2}f(t+m)^2$$ ……………$$(1)$$
Also $$V=0+ht=0+f(t+m)$$
$$\Rightarrow t+m=\dfrac{ht}{f}$$
From equation $$(1)$$,
$$\dfrac{1}{2}ht^2+n=\dfrac{1}{2}f\left(\dfrac{ht}{f}\right)^2$$
$$\Rightarrow t^2=\dfrac{2nf}{h(h-f)}$$
Also,
$$ht=f(t+m)$$
$$\Rightarrow t^2=\dfrac{m^2f^2}{(h-f)^2}$$
$$\therefore \dfrac{2nf}{h(h-f)}=\dfrac{m^2f^2}{(h-f)^2}$$
$$\Rightarrow 2n=\dfrac{m^2fh}{h-f}$$
$$\Rightarrow n(h-f)=\dfrac{1}{2}fhm^2$$.
Question 10
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If $$x= a ( cos\theta+\theta sin\theta)$$ and $$y= a( sin\theta -\theta cos \theta)$$ and $$\theta$$ increases at uniform rate $$\omega$$. The velocity of particle is

A
$$a\omega$$

B
$$\frac{a^2\theta}{\omega}$$

C
$$\frac{a\theta}{\omega}$$

D
$$a \theta\omega$$

Solution