Motion in A Straight Line Test

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Motion in A Straight Line Test - 23

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Question 1
1 / -0

A ball is projected vertically up with a velocity of $$40$$ $$m/s$$. At the same time another ball is dropped from a height of $$100$$ m. The magnitudes of their velocities are equal after a time of $$(g= 10m/s^{2})$$

A
2 sec

B
1 sec

C
3 sec

D
4 sec

Solution

$$v_A = 40 - gt$$
$$v_B = gt$$
$$|v_A| = |v_B|$$
$$\Rightarrow 40 - gt = gt$$
$$t = \dfrac {40}{2 \times 10} = 2\ sec$$
Question 2
1 / -0

A car, starting from rest, accelerates at the rate $$f$$ through a distance $$S$$, then continues at constant speed for time $$t$$ and then decelerate at the rate $$\frac{f}{2}$$ to come to rest. If the total distance travelled is $$15S$$, then

A
$$S=ft$$

B
$$S=\dfrac{1}{6}ft^{2}$$

C
$$S=\dfrac{1}{72}ft^{2}$$

D
$$S=\dfrac{1}{4}ft^{2}$$

Solution

$$\displaystyle S_1 = \dfrac{1}{2}ft_{1}^{2}=S$$ since given $$S_1 = S$$
$$S_2 =( ft_1)t_2$$
Since acceleration for $$t_3$$ is $$\dfrac{-f}{2}$$, $$t_3=2t_1$$
$$S_3 =( ft_1)t_3 - \dfrac{1}{2}(\dfrac{f}{2})t_3 ^2$$
$$ \therefore S_3 = ( ft_1)(2t_1) - \dfrac{1}{2}(\dfrac{f}{2})(2t_1)^2= 2ft_1^2=2S_1=2S$$
Given, $$S_1 + S_2 +S_3 = 15S$$
$$\Rightarrow S + S_2 + 2S =15S$$
$$\Rightarrow S_2 = 12S$$
$$\Rightarrow ft_1t_2 = 12S= 12 \times \dfrac{1}{2} \times ft_1^2$$
$$ \Rightarrow t_2 = 6t_1$$
$$S =\dfrac{1}{2}ft_1 ^2 = \dfrac{1}{72}ft_2 ^2 = \dfrac{1}{72}ft ^2$$
(since given $$t_2 = t$$)
Question 3
1 / -0

A point moves with uniform acceleration. Let $$v_{1}$$, $$v_{2}$$, $$v_{3}$$ denote the average velocities in three successive intervals of time $$t_{1}$$, $$t_{2}$$, $$t_{3}$$. Correct relation among the following is

A
$$(v_{1}-v_{2})$$ : $$(v_{2}-v_{3})$$ = $$(t_{1}-t_{2})$$ : $$(t_{2}-t_{3})$$

B
$$(v_{1}-v_{2})$$ : $$(v_{2}-v_{3})$$ = $$(t_{1}+t_{2})$$ : $$(t_{2}+t_{3})$$

C
$$(v_{1}-v_{2})$$ : $$(v_{2}-v_{3})$$ = $$(t_{1}-t_{2})$$ : $$(t_{2}+t_{3})$$

D
$$(v_{1}-v_{2})$$ : $$(v_{2}-v_{3})$$ = $$(t_{1}+t_{2})$$ : $$(t_{2}-t_{3})$$

Solution

$${ v }_{ avg }=\dfrac { ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 } }{ t } =u+\dfrac { 1 }{ 2 } at$$

$${V}_{1}={U}+\dfrac{a{t}_{1}}{2}$$

$${V}_{2}={U}_{1}+\dfrac{a{t}_{2}}{2}$$

$${V}_{3}={U}_{2}+\dfrac{a{t}_{3}}{2}$$

so we can find,

$$({V}_{1}-{V}_{2}):({V}_{2}-{V}_{3})=({t}_{1}-{t}_{2}):({t}_{2}-{t}_{3})$$

Alternatively,

Since acceleration is constant, acceleration in the interval $$t_1 -t_2$$ will be same as acceleration in the interval $$t_2 -t_3$$
Hence,
$$({V}_{1}-{V}_{2}):({V}_{2}-{V}_{3})=({t}_{1}-{t}_{2}):({t}_{2}-{t}_{3})$$
Question 4
1 / -0

A man in a lift ascending with an upward acceleration throws a ball vertically upwards and catches it after $$t_{1}$$ second. Later when the lift is descending with the same acceleration, the man throws the ball up again with same velocity and catches it after $$t_{2}$$ second.
A) The acceleration of the elevator is $$g\dfrac{(t_{2}-t_{1})}{(t_{1}+t_{2})}$$
B) The velocity of projection of the ball relative to elevator is $$\dfrac{t_{2}t_{1} g}{t_{1}+t_{2}}$$.
We can conclude that:

A
only $$A$$ is true

B
only $$B$$ is true

C
Both $$A$$ and $$B$$ is true

D
Both $$A$$ and $$B$$ are false

Solution

We know that
$$ { t }_{ 1 }=\dfrac { 2u }{ g+a } \quad \& \quad { t }_{ 2 }=\dfrac { 2u }{ g-a }$$

We get,
$$2u={ t }_{ 1 }(g+a)={ t }_{ 2 }(g-a)\\ a({ t }_{ 1 }+{ t }_{ 2 })=g({ t }_{ 2 }-{ t }_{ 2 })\\ a=\dfrac { g({ t }_{ 2 }-{ t }_{ 1 }) }{ { t }_{ 1 }+{ t }_{ 2 } } $$

Now, the velocity of projection of the ball relative to elevator is u, We have,
$$g +a = \displaystyle \frac{2u}{t_1}$$
$$g - a = \displaystyle \frac{2u}{t_2}$$

adding both,
$$2g = 2u \displaystyle(\frac{1}{t_1} + \frac{1}{t_1})$$

$$\therefore u = \displaystyle \frac{t_1t_2g}{t_1+t_2}$$
Question 5
1 / -0

A body is thrown vertically upwards with an initial velocity $$u$$ reaches a maximum height in $$6s$$. The ratio of the distance travelled by the body in the first second to the seventh second is

A
1:1

B
11:1

C
1:2

D
1:11

Solution

height in the first second,
$$h_{1}=u-\displaystyle \frac{g}{2}$$
hight covered in the first second of downward journey
$$h_{2}=\displaystyle \frac{g}{2}$$
given $$t_{a}=\displaystyle \frac{u}{g}=6; u=6g$$

$$v = 0 $$ at max height
$$v^2 = u^2 - 2gs | t = \dfrac {u}{g}$$
$$\Rightarrow u^2 = 2gH_{max} | u = gt$$

$$u = \sqrt {2gH_{max}} | u = 6g - (0)$$
$$s_1 = u(1) - \dfrac {1}{2}g(1)^2 = u - \dfrac {g}{2} \ \ \ ..... (1)$$
$$s_7 = [u(7)-\dfrac {1}{2}g(7)^2] - [u(6) - \dfrac {1}{2}g(6)^2]$$
$$= u -\dfrac {1}{2}g(13) \ \ \ ..... (2)$$
$$\dfrac {s_1}{s_7} = \dfrac {u-\dfrac {g}{2}}{u - \dfrac {g}{2}(13)} = \dfrac {6g - \dfrac {g}{2}}{6g - \dfrac {13g}{2}} = \dfrac {11}{-1} \ \ \ ..... (3)$$
So, $$\dfrac {s_1}{s_7} = \dfrac {11}{1}$$
Question 6
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A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by $$x_{1}(t)$$ after time t and that of the second body by $$ x_{2} (t)$$ after the same time interval. Which of the following graphs correctly describes $$( x_{1} - x_{2})$$ as a function of time t ?

A

B

C

D

Solution

Relative distance at any point
$$S_{1} (t) = 1/2 at^{2}$$
$$S_{2} (t) = Vt$$
$$\Rightarrow (S_{1} - S_{2}) =1/2 at^{2}-vt$$
Curve in D seems to be correct answer.
Question 7
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An object falls from a bridge that is $$45\ m$$ above water. It falls directly into a small boat moving with constant velocity that is $$12\ m$$ from the point of impact when the object was released. The speed of the boat is

A
$$3\;m/s$$

B
$$4\;m/s$$

C
$$5\;m/s$$

D
$$6\;m/s$$

Solution

Boat has covered $$12\ m$$ in the same time as taken by ball A to fall down, $$t = \dfrac {12}{v}$$
Time taken by A $$\Rightarrow 45 = \dfrac {1}{2}gt^2$$

$$t = \sqrt {\dfrac {90}{g}}=3 sec $$

Now, $$\dfrac {12}{v} = 3 $$

$$\Rightarrow v = 4 \ m/s$$
Question 8
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A body, projected vertically upwards, crosses a point twice its journey at a height $$h$$ just after $$t_{1}$$ and $$t_{2}$$ seconds. Maximum height reached by the body is

A
$$\dfrac{g}{4}(t_{1}+t_{2})^{2}$$

B
$$g\left [\dfrac{t_{1}+t_{2}}{4}\right ]^{2}$$

C
$$2g\left [\dfrac{t_{1}+t_{2}}{4}\right ]^{2}$$

D
$$\dfrac{g}{4} (t_{1}t_{2})$$

Solution

$$s_A = ut - \dfrac {1}{2}gt^2 (at \;t = t_1, t_2)$$
So, $$s_A = ut_1 - \dfrac {1}{2}gt_1^2 - (1)$$
$$s_A = ut_2 - \dfrac {1}{2} gt_2^2 - (2)$$

Equating $$(1) and (2)$$
$$\Rightarrow u (t_1 - t_2) = \dfrac {1}{2} g (t_1^2 - t_2^2)$$
$$\Rightarrow u = g \dfrac {(t_1 + t_2)}{2}$$
$$Max\ ht = \dfrac {u^2}{2g} = \dfrac {g^2 (t_1 + t_2)^2}{4 \times 2g} = 2g (\dfrac {t_1 + t_2}{4})^2$$
Question 9
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A train is moving forward at a velocity of $$2.0$$ $$m/s$$. At the instant the train begins to accelerate at $$0.80 m/s^{2}$$, a passenger drops a coin which takes $$0.50\ s$$ to fall on the floor. Relative to a spot on the floor directly under the coin at release, the coin lands

A
$$1.1$$ m towards the rear of the train

B
$$1.0$$ m towards the rear of the train

C
$$0.10$$ m towards the rear of the train

D
$$0.90$$ m towards the front of the train

Solution

We have to solve this problem in reference to frame of train
So, Relative velocity of coin w.r.t train $$= 0 m/s$$
Relative acceleration of coin w.r.t train = -0.8 $$m/{ { s }^{ 2 } }$$
Here, (-) indicates that direction opposite to train initial direction
Then, by using 2nd equation of motion
$$ S=ut+$$ $$\frac { 1 }{ 2 } a{ t }^{ 2 }$$
$$ S= 0+$$ $$\frac { 1 }{ 2 } \times (-0.8)\times { (0.5) }^{ 2 }$$
$$ S= -0.10 m$$
Therefore, the final answer is the coin moves 0.01 m towards the rear of the train.
Question 10
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A ball is thrown vertically upwards with a velocity $$u$$ from the balloon descending with velocity $$v$$. The ball will pass by the balloon after time

A
$$\dfrac{u-\nu}{2g}$$

B
$$\dfrac{2(u+\nu)}{g}$$

C
$$\dfrac{2u-\nu}{g}$$

D
$$\dfrac{2u+\nu}{g}$$

Solution

$${ \vec { v } }_{ BB }=$$ Relative velocity of ball with respect to balloon $$=\vec { u } +\vec { v }$$

$$0=-\left( u+v \right) +gt$$ $$\implies t=\displaystyle\dfrac { u+v }{ g }$$

Total time $$=\displaystyle\dfrac { 2\left( u+v \right) }{ g }$$