Motion in A Straight Line Test

Result

Motion in A Straight Line Test - 42

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

An astronaut jumps from an airplane. After he had fallen $$40 m$$, then his parachute opens. Now he falls with a retardation of $$2 m/s^2$$ and reaches the earth with a velocity of $$3.0 m/s$$. What was the height of the aeroplane? (in m)

A
$$185$$

B
$$190$$

C
$$234$$

D
$$200$$

Solution

Velocity after falling through 40m=$$\sqrt{2gh}=\sqrt{2\times 9. 81 \times 40}=28m/s$$
Distance travelled after opening the parachute=$$(28^2-3^2)/2\times 2=194$$
So, height of the aeroplane$$=194+40=234m$$

Hence none of the above options are correct.
Question 2
1 / -0

A car starts from rest and accelerates uniformly to a speed of $$ 180 \,km \,h^{-1} $$ in $$10$$ sec. The distance covered by the car in the time interval is :

A
$$200\, m$$

B
$$300\, m$$

C
$$500 \,m$$

D
$$250\, m$$

Solution

$$u = 0$$,
$$v = 180$$
$$180km/h = 50 m/s$$
Time taken, $$t = 10 s$$
$$ a = \dfrac {v-u}{t} = \dfrac {60}{10} = 6 ms^{-2} $$
Distance covered
$$ s = ut + \dfrac {1}{2} at^2 $$
$$ = 0 + \dfrac {1}{2} \times 5 \times ( 10 )^2 $$
$$ = \dfrac {500}{2} \Rightarrow = 250 m $$
Question 3
1 / -0

Two stones of mass $$m_1$$ and $$m_2$$ (such that $$m_1$$ and $$m_2$$ ) are dropped $$\triangle$$ t time apart from the same height towards the ground. At a later time t the difference in their speed is $$\triangle $$ V and their mutual separation is $$\triangle$$ S . While both stones are in flight

A
$$\triangle$$ V decreases with time and $$\triangle$$ S increases with time.

B
$$\triangle$$ V and $$\triangle$$ S increases with time.

C
$$\triangle$$ V remains constant time and $$\triangle$$ S decreases with time.

D
$$\triangle$$ V remains constant time and $$\triangle$$ S increases with time.

Solution

$$\Delta V=g\Delta t$$
Since $$\Delta t$$ is unchanged, $$\Delta V$$ remains same.
After some time t,
$$\Delta S=\frac{gt^2}{2}-\frac{g(t-\Delta t)^2}{2}=\frac{1}{2}(2gt\Delta t+(\Delta t)^2)$$
$$\Delta S \propto t$$ so $$\Delta S$$ increases with time.
Question 4
1 / -0

A particle moves along the x-axis with a position given by the equation $$x = 5 + 3t$$, where $$x$$ is in meters, and $$t$$ is in seconds. The positive direction is east. Which of the following statements about the particle is false?

A
The particle is east of the origin at $$t = 0$$

B
The particle is at rest at $$t = 0$$

C
The particle's velocity is constant

D
The particle's acceleration is constant

Solution

Given position vector is: $$x = 5 + 3t$$
Hence, at $$t = 0, \quad x = 5\ m$$ i.e., particle is east of origin. Option A is correct.

Comparing with second equation of motion,
$$s = ut + \cfrac{1}{2}at^2$$
$$u = 3\ m/s$$
Option B is false

$$a = 0$$ Thus, option D is true.
Hence, particle's velocity is constant. Option C is true.
Question 5
1 / -0

A car moving with a velocity of $$20{ ms }^{ -1 }$$ is stopped in a distance of $$40 m$$. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is

A
$$640 m$$

B
$$320 m$$

C
$$1280 m$$

D
$$160 m$$

Solution

Initial speed of the car is $$u$$.
Final speed of the car $$v=0$$
Let the retardation of the car be $$a$$.
Using $$v^2 - u^2 = 2aS$$
Or $$0- u^2 = 2aS$$
$$\implies$$ $$S\propto u^2$$ (for same $$a$$)
Given : $$S = 40 m$$ $$u' = 2u$$
$$\therefore$$ $$\dfrac{S'}{S} = \dfrac{(u')^2}{u^2} =2^2$$
Or $$\dfrac{S'}{40} = 4$$
$$\implies$$ $$S' = 160 m$$
Question 6
1 / -0

A man weighing 80 kg is standing on a trolley weighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley along the rails at speed 1 m/s. Then after 4 s, his displacement relative to ground will be :

A
4.5 m

B
5 m

C
8 m

D
3.2 m

Solution

In $$1s$$, man has walked, $$4m/s\times1s = 4m$$, with respect to the trolley.
As position of the CoM stays same, let the distance moved by trolley be $$x$$ in direction opposite to that of man.
So net diplacement of man is $$(4-x)m$$

Thus, $$M_{man}\times(4-x) = M_{trolley}x$$
$$\implies 80(4-x) = 320x\implies x=320/400=0.8m$$

So net displacement of man in $$1s$$ is, $$4-0.8 = 3.2m$$
Option D is correct.
Question 7
1 / -0

A body travelling along a straight line one-third of the total distance with a velocity $$4m/s$$. The remaining part of the distance was covered with a velocity $$2m/s$$ for the first half of the remaining journey and with a velocity $$6m/s$$ for the another half of the remaining journey. The average velocity is

A
$$5m/s$$

B
$$4m/s$$

C
$$4.5m/s$$

D
$$3.5m/s$$

Solution

Let total distance d
Let time covered for first one third distance be$$t_1$$
Then $$t_1=\large\frac{d/3}{4}=\large\frac{d}{12}$$
$$t_2$$ be time for next two journeys
$$d-\frac{d}{3}=\frac{2d}{3}=2t_2+6 t_2=8t_2$$
$$t_2=\large\frac{d}{12}$$
$$Average\;velocity=\large\frac{Total\;distane}{Total\;time}=\frac{d}{t_1+2t_2}=\large\frac{d}{\Large\frac{d}{12}+\frac{2d}{12}}=4m/s$$
Question 8
1 / -0

A particle moving in one dimension with a constant acceleration of $$2 {m}/{{s}^{2}}$$ is observed to cover a distance of $$5 m$$ during a particular interval of $$1 s$$. The distance covered by the particle in the next $$1 s$$ interval is in metre

A
$$5$$

B
$$6$$

C
$$7$$

D
$$10$$

Solution

The distance cover 5 m in time interval 1 s means its speed is $$5 m/s$$.
Using, $$v=at$$, $$t=v/a=5/2 s$$
For next interval, $$t_1=t+1=5/2+1=7/2 s$$
For after time $$t_1$$, the speed will be $$v_1=at_1=2(7/2)=7 m/s$$
Thus, distance covered by next 1 s interval will be $$7 m$$.
Question 9
1 / -0

In a car race, car $$A$$ takes a time $$t$$ less than car $$B$$ at the finish and passes the finishing point with speed $$v$$ more than that of the car $$B$$. Assuming that both the cars starts from rest and travel with constant accelerations $${ a }_{ 1 }$$ and $${ a }_{ 2 }$$ respectively. So, the value of $$v$$ will be :

A
$$\left( \sqrt { { { a }_{ 1 } }/{ { a }_{ 2 } } } \right) t$$

B
$$\left( \sqrt { { { a }_{ 2 } }/{ { a }_{ 1 } } } \right) t$$

C
$$\left( { a }_{ 1 }\sqrt { { a }_{ 2 } } \right) t$$

D
$$\left( \sqrt { { a }_{ 1 }{ a }_{ 2 } } \right) t$$

Solution

Consider that $$A$$ takes $${ t }_{ 1 }$$ second, then according to the given problem, $$B$$ will take $$\left( { t }_{ 1 }+t \right) $$ seconds. Further let $${ v }_{ 1 }$$ be the velocity of $$B$$ at finishing point, then velocity of $$A$$ will be $$\left( { v }_{ 1 }+v \right)$$. Writing equations of motion for $$A$$ and $$B$$
$${ v }_{ 1 }+v={ a }_{ 1 }{ t }_{ 1 }$$ ....(i)
$${ v }_{ 1 }={ a }_{ 2 }\left( { t }_{ 2 }+t \right) $$ ....(ii)
From equations (i) and (ii), we get
$$v=\left( { a }_{ 1 }-{ a }_{ 2 } \right) { t }_{ 1 }-{ a }_{ 2 }t$$ .....(iii)
Total distance travelled by both the cars is equal
$${ S }_{ A }={ S }_{ B }$$
$$\Rightarrow \dfrac { 1 }{ 2 } { a }_{ 1 }{ t }_{ 1 }^{ 2 }=\dfrac { 1 }{ 2 } { a }_{ 2 }{ \left( { t }_{ 1 }+t \right) }^{ 2 }\Rightarrow { t }_{ 1 }=\dfrac { \sqrt { { a }_{ 2 } } t }{ \sqrt { { a }_{ 1 } } -\sqrt { { a }_{ 2 } } }$$
Substituting this value of $${ t }_{ 1 }$$ in equation (iii), we get
$$v=\left( \sqrt { { a }_{ 1 }{ a }_{ 2 } } \right) t$$
Question 10
1 / -0

A body freely falling from the rest has a velocity v after his falls through a height h. The distance, it has to fall down further for its velocity to become double, is :

A
4 h

B
3h

C
h

D
16h

Solution

$$\textbf{Hint:}$$ Apply third equation of motion.

$$\textbf{Step 1-Use third equation of motion }$$
If a body falls freely so its initial velocity is $$u=0$$
Acceleration is $$a=-g$$ and displacement $$s=-h$$
After travelling distance h velocity becomes v.

So $$v^2=u^2+2as$$
$$\Rightarrow v^2=0+2(-g)(-h)=2gh$$.......(1)

$$\textbf{Step 2-Calculate distance when velocity becomes double}$$
After travelling distance $$h_1$$,velocity becomes double i.e.$$2v$$.

Again apply third equation of motion
$$(2v)^2=v^2+2(-g)(-h_1)$$
$$\Rightarrow 4v^2=v^2+2gh_1$$.......(2)

$$\textbf{Step 3-Solve above equations to obtain answer}$$
Put value from equation (1)
$$4(2gh)=(2gh)+2gh_1$$
$$\Rightarrow 2gh_1=6gh$$
$$\Rightarrow h_1=3h$$

Hence the correct answer is option (B)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 42

Result

Motion in A Straight Line Test - 42

Score

-

Rank

-

SHARING IS CARING

Question 1

$$185$$

$$190$$

$$234$$

$$200$$

Solution

Question 2

$$200\, m$$

$$300\, m$$

$$500 \,m$$

$$250\, m$$

Solution

Question 3

$$\triangle$$ V decreases with time and $$\triangle$$ S increases with time.

$$\triangle$$ V and $$\triangle$$ S increases with time.

$$\triangle$$ V remains constant time and $$\triangle$$ S decreases with time.

$$\triangle$$ V remains constant time and $$\triangle$$ S increases with time.

Solution

Question 4

The particle is east of the origin at $$t = 0$$

The particle is at rest at $$t = 0$$

The particle's velocity is constant

The particle's acceleration is constant

Solution

Question 5

$$640 m$$

$$320 m$$

$$1280 m$$

$$160 m$$

Solution

Question 6

4.5 m

5 m

8 m

3.2 m

Solution

Question 7

$$5m/s$$

$$4m/s$$

$$4.5m/s$$

$$3.5m/s$$

Solution

Question 8

$$5$$

$$6$$

$$7$$

$$10$$

Solution

Question 9

$$\left( \sqrt { { { a }_{ 1 } }/{ { a }_{ 2 } } } \right) t$$

$$\left( \sqrt { { { a }_{ 2 } }/{ { a }_{ 1 } } } \right) t$$

$$\left( { a }_{ 1 }\sqrt { { a }_{ 2 } } \right) t$$

$$\left( \sqrt { { a }_{ 1 }{ a }_{ 2 } } \right) t$$

Solution

Question 10

4 h

3h

h

16h

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.