Motion in A Straight Line Test

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Motion in A Straight Line Test - 64

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Question 1
1 / -0

A particle travels so that its acceleration is given by $$\vec{a}=5\,cos\,t\hat {i}-3\,sin\,t\hat {j}$$. If the particle is located at $$(-3,2)$$ at time $$t=0$$ and is moving with a velocity given by $$(-3\hat {i}+2\hat {j})$$. Find
(i) the velocity $$\begin{bmatrix}\vec{v}=\int \vec{a}.dt\end{bmatrix}\:time\:t$$
(ii) the position vector $$[\vec{r}=\int \vec{v}.dt]$$ of the particle at time $$(t>0)$$.

A
(i)$$\;\vec{v}=(5sin\;t-3)\hat {i}+(3cos\;t-1)\hat {j}$$, (ii)$$\;\vec{s}=(2-5cos\;t-3t)\hat {i}+(2+3sin\;t-t)\hat {j}$$

B
(i)$$\;\vec{v}=(3sin\;t-3)\hat {i}+(3cos\;t-1)\hat {j}$$, (ii)$$\;\vec{s}=(2-5cos\;t-3t)\hat {i}+(2+3sin\;t-t)\hat {j}$$

C
(i)$$\;\vec{v}=(4sin\;t-3)\hat {i}+(3cos\;t-1)\hat {j}$$, (ii)$$\;\vec{s}=(2-5cos\;t-3t)\hat {i}+(2+3sin\;t-t)\hat {j}$$

D
None of these

Solution

$$\vec{a}=5\,cos\,t\hat {i}-3\,sin\,t\hat {j}$$
$$\Rightarrow \int d\vec{v}=\int 5\,cos\,t\;dt\hat {i}-\int 3\,sin\,t\;dt\hat {j}$$
Therefore $$\underset{-3}{\overset{v}{\int}}dv_{x}=\underset{0}{\overset{t}{\int}}5\,cos\,t\;dt\Rightarrow v_x=5\,sin\,t-3$$
$$\displaystyle\frac{dx}{dt}=(5sint-3)\Rightarrow \underset{-3 }{\overset{x}{\int}}dx=\underset {0}{\overset{t}{\int}}(5\,sin\,t-3)dt$$
$$x+3=5-5\,cos\,t-3t\Rightarrow x=2-5\,cos\,t-3t$$
Similarly,
$$\underset{2}{\overset{v}{\int}}dv_{y}=-\underset{0}{\overset{t}{\int}}3\,sin\,t\,dt$$
$$\Rightarrow v_y-2=3(cos\,t-1)\Rightarrow v_y=3\,cos\,t-1$$
$$\Rightarrow \underset{2}{\overset{y}{\int}}dy=\underset{0}{\overset{t}{\int}}(3\,cos\,t-1)dt$$
$$\Rightarrow y-2=3\;sin\,t-t\Rightarrow y=2+3\,sin\,t-t$$
Thus, $$\vec{v}=(5\,sin\,t-3)\hat {i}+(3\,cos\,t-1)\hat {j}$$
and $$\vec{s}=(2-5\,cos\,t-3\,t)\hat {i}+(2+3\,sin\,t-t)\hat {j}$$
Question 2
1 / -0

Two balls were thrown vertically upwards with different velocities, what is the shape of the graph between distance between the balls and time before either of the two collide with ground?

A
Straight line passing through origin

B
Parabola

C
Circle

D
None of the above

Solution

Let the initial velocity of the balls be $$u_1$$ and $$u_2$$ (upwards) respectively.
Also the acceleration of the balls $$a_1 = -g$$ $$a_2 =-g$$
$$\therefore$$ Relative acceleration of ball 1 w.r.t 2 $$a_{12} =a_1 -a_2 = 0$$
Initial velocity of 1 w.r.t 2 $$u_{12} =u_1 - u_2 $$ (upwards)
Using $$S_{12} = u_{12} t - \dfrac{a_{12} t^2}{2}$$
$$\therefore$$ Distance between them $$S_{12} = (u_1 - u_2) t$$
Thus S-t graph will be a straight line having a slope $$(u_1 - u_2)$$ passing through origin.
Question 3
1 / -0

A thief is running away on a straight road with a speed of $$9 \ ms^{-1}$$. A police man chases him on a jeep moving at a speed of $$10 \ ms^{-1}$$ If the instantaneous separation of the jeep from the motorcycle is 100 m, how long will it take. for the police man to catch the thief?

A
1s

B
19s

C
90s

D
100 s

Solution

Police man is advancing with a speed of 1 m/s (difference between the speeds)
Distance between the two= 100 m
Hence, time = 100/1= 100 s
Question 4
1 / -0

A stone is thrown upwards from a tower with a velocity $$50\, ms^{-1}$$. Another stone is simultaneously thrown downwards from the same location with a velocity $$50\,ms^{-1}$$ . When the first stone is at the highest point, the relative velocity of the second stone with respect to the first stone is (assume that second stone has not yet, reached the ground

A
Zero

B
$$50\, ms^{-1}$$

C
$$100\, ms^{-1}$$

D
$$150\, ms^{-1}$$

Solution

Initially one stone is thrown up with a velocity of $$50ms^{-1}$$
Time it takes to reach stationary.
v =0, u =50, a = -10,
v = u + at
Solving we get 5s.
Meanwhile in 5seconds the other rock initially moving down with u $$ = 50ms^{-1}$$ has now attained a velocity v.
$$a = +10 ms^{-2}$$
$$v = u + at = 50 + 5(10) = 100 ms^{-1}$$
Hence one rock has a velocity of 0, and the other has a velocity of $$100ms^{-1}$$
Hence relative velocity is $$100ms^{-1}$$ option C is correct.
Question 5
1 / -0

A ballon starts rising from the ground with an acceleration of $$1.25 \ m/s^{2}$$. After $$8\ s$$ a stone is released from the balloon. The stone will (take $$g=10\ m/s^{2}$$)

A
have a displacement of 50 m

B
cover a distance of 40 m in reaching the ground

C
reach the ground in 4 s

D
reach the ground in 16 s

Solution

at t=8sec. height of ballon

$$\displaystyle h=0+\frac{1}{2}\times 1.25\times (8)^{2}=40m$$

and velocity of ballon $$v_{b}=0+1.25\times 8=10m/s (upward)$$

And as the stone is released. It has the same initial velocity i.e 10m/sec (upward) hence it will go up by 5m and come back 5m and then fall down a distance 40m downwards so the total distance covered by stone is 40m downwards. Now if the time of journey is t then

$$\displaystyle h=ut+\frac{1}{2}at^{2}\Rightarrow +40=-10t+\frac{1}{2}\times 10\times t^{2}$$
$$\Rightarrow t^{2}-2t-8=0\Rightarrow (t-4)(t+2)=0$$
$$\Rightarrow t=4s.(t=-2\;not\;possible)$$
Question 6
1 / -0

Which of the following is not an example of acceleration?

A
A person jogging at $$3\ m/s$$ along a winding path

B
A car stopping at a stop sign

C
A cheetah running $$27\ m/s$$ east

D
A plane taking off

Solution
Question 7
1 / -0

Consider a train which can accelerate with an acceleration of $$20 {cm}/{{s}^{2}}$$ and slow down with deceleration of $$100 {cm}/{{s}^{2}}$$. Find the minimum time for the train to travel between the stations $$2.7 km$$ apart.

A
0

B
90

C
100

D
180

Solution
Question 8
1 / -0

An aeroplane flies along a straight line from A to B with air speed $$V$$ and back again with the same air speed.If the distance between A and B is $$l$$ and a steady wind blows perpendicular to AB with speed $$u$$, the total time taken for the round trip is

A
$$\dfrac{2l}{\sqrt{V^{2}-u^{2}}}$$

B
$$\dfrac{2l}{\sqrt{V^{2}+u^{2}}}$$

C
$$\dfrac{l}{\sqrt{V^{2}-u^{2}}}$$

D
$$\dfrac{3l}{\sqrt{V^{2}-u^{2}}}$$

Solution

If the plane has to fly in straight line then he must fly at some angle theta with the AB. And component of velocity of plane in direction of air flow with respect to ground should be zero. i.e. $$u=vSin\theta$$. From here $$Sin\theta=\frac{u}{v}$$
And component of velocity of plane which will help plane to cover the distance $$l$$ is $$vCos\theta$$
So time taken to cover distance $$l$$ is given by $$\frac{l}{vCos\theta}$$ and $$Cos\theta=\sqrt{1-(\frac{u}{v})^2}$$.
So time taken from A to B is $$\frac{l}{\sqrt{v^2-u^2}}$$ same will be case while returning so total time will be $$\frac{2l}{\sqrt{v^2-u^2}}$$
So best option is A.
Question 9
1 / -0

A person drops a metallic sphere from the top of a tower of height 125 m. Another person at a distance of 80 m from the foot of the tower hears the sound of the sphere hitting the ground after a time interval of 5.25 s. Find the velocity of the sound in air.($$g = 10 {ms}^{-2}$$)

A
$$140{ms}^{-1}$$

B
$$280{ms}^{-1}$$

C
$$320{ms}^{-1}$$

D
$$460{ms}^{-1}$$

Solution
Question 10
1 / -0

At a certain acceleration, a racing car takes $$5 s$$ to double its speed while moving at $$2 m{s}^{-1}$$. If now, it moves at $$4 m{s}^{-1}$$, at the same acceleration, what time will it take to double its speed?

A
$$5s$$

B
$$10s$$

C
$$15s$$

D
None of these

Solution

Let the acceleration be $$a$$.
$$v=u+at$$
$$4=2+(a\times 5)$$
$$a=\dfrac{2}{5}ms^{-2}$$
Let the time required be $$T$$.
$$\therefore 8=4+(\dfrac{2}{5}\times T)$$
$$4=\dfrac{2}{5}\times T$$
$$T=10\ s$$

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Motion in A Straight Line Test - 64

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Motion in A Straight Line Test - 64

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SHARING IS CARING

Question 1

(i)$$\;\vec{v}=(5sin\;t-3)\hat {i}+(3cos\;t-1)\hat {j}$$, (ii)$$\;\vec{s}=(2-5cos\;t-3t)\hat {i}+(2+3sin\;t-t)\hat {j}$$

(i)$$\;\vec{v}=(3sin\;t-3)\hat {i}+(3cos\;t-1)\hat {j}$$, (ii)$$\;\vec{s}=(2-5cos\;t-3t)\hat {i}+(2+3sin\;t-t)\hat {j}$$

(i)$$\;\vec{v}=(4sin\;t-3)\hat {i}+(3cos\;t-1)\hat {j}$$, (ii)$$\;\vec{s}=(2-5cos\;t-3t)\hat {i}+(2+3sin\;t-t)\hat {j}$$

None of these

Solution

Question 2

Straight line passing through origin

Parabola

Circle

None of the above

Solution

Question 3

1s

19s

90s

100 s

Solution

Question 4

Zero

$$50\, ms^{-1}$$

$$100\, ms^{-1}$$

$$150\, ms^{-1}$$

Solution

Question 5

have a displacement of 50 m

cover a distance of 40 m in reaching the ground

reach the ground in 4 s

reach the ground in 16 s

Solution

Question 6

A person jogging at $$3\ m/s$$ along a winding path

A car stopping at a stop sign

A cheetah running $$27\ m/s$$ east

A plane taking off

Solution

Question 7

0

90

100

180

Solution

Question 8

$$\dfrac{2l}{\sqrt{V^{2}-u^{2}}}$$

$$\dfrac{2l}{\sqrt{V^{2}+u^{2}}}$$

$$\dfrac{l}{\sqrt{V^{2}-u^{2}}}$$

$$\dfrac{3l}{\sqrt{V^{2}-u^{2}}}$$

Solution

Question 9

$$140{ms}^{-1}$$

$$280{ms}^{-1}$$

$$320{ms}^{-1}$$

$$460{ms}^{-1}$$

Solution

Question 10

$$5s$$

$$10s$$

$$15s$$

None of these

Solution

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