Motion Test

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Motion Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

Odometer is to mileage as compass is to

A
speed

B
hiking

C
needle

D
direction

Solution

An odometer is an instrument used to measure mileage. A compass is an instrument used to determine direction. Choices a, b, and c are incorrect because none is an instrument
Question 2
1 / -0

The odometer of a car reads $$57321.0 km$$ when the clock shows the time $$08:30$$ AM. What is the distance moved by the car, if at $$08:50 AM$$, the odometer reading has changed to $$57336.0 km$$?

A
$$75\ km$$

B
$$50\ km$$

C
$$15\ km$$

D
$$10\ km$$

Solution

An odometer is a device that gives us the distance covered by a car.
Distance covered by the car in the given time is $$d = 57336.0 - 57321 = 15.0 \ km$$
Question 3
1 / -0

The velocity of the object over the interval (2,4) is :

A
10 m/s

B
5 m/s

C
20 m/s

D
15 m/s

Solution

Velocity is defined as the rate of change of position with respect to time.
$$v=\dfrac{(s_2-s_1)}{(t_2-t_1)}$$

$$v=\dfrac{(20-10)}{(4-2)}$$

$$v=\dfrac{10}{2}$$

$$v=5\ m/s$$
Question 4
1 / -0

The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on the planet B. A man jumps to a height of 2m on the surface A. What is the height of jump by the same person on the plane B?

A
6 m

B
2/9 m

C
2/3 m

D
18 m

Solution

It is given that acceleration due to gravity on plane. A is 9 times the acceleration due to gravity on planet B i.e.,
$$ g_A = 9g_B $$ ...(i)
From third equation of motion
$$ v^2 = 2gh $$ ...(ii)
At planet $$ A, h_A = \dfrac {v_2}{2g_A} $$ ...(ii)
At planet $$ A, h_B = \dfrac {v_2}{2g_B} $$ ...(iii)
Dividing Eq. (ii) by Eq. (iii) we have
$$ \dfrac {h_A}{h_B} = \dfrac {9g_B} { 9g_A} $$
From Eq. (i), $$ g_A = 9 g_B $$
$$ \therefore \dfrac {h_A}{h_B} = \dfrac {9g_B} { 9g_A} = \dfrac {1}{9} $$
or $$ h_B = 9 h_B = 9 \times 2 = 18 m ( \therefore h_A = 2m ) $$
Question 5
1 / -0

An object is thrown vertically upward with a speed of $$30m/s$$. The velocity of the object half a second before it reaches the maximum height is

A
$$4.9m/s$$

B
$$9.8m/s$$

C
$$19.6m/s$$

D
$$25.1m/s$$

Solution

Velocity half second before maximum height $$=$$ Velocity half second after maximum height (Return journey)
For downward journey :
Initial velocity $$u = 0 \ m/s$$
We have $$g = 9.8 \ m/s^2$$
Time $$t = \dfrac{1}{2} \ s$$
Thus velocity after half second $$v = u+gt$$
$$v=0+9.8\times \dfrac{1}{2}=4.9m/s$$
Question 6
1 / -0

Two cars started moving with initial velocities $$u$$ and $$2u$$. For the same deceleration, their respective stopping distances are in the ratio

A
$$1 : 1$$

B
$$1 : 2$$

C
$$1 : 4$$

D
$$2 : 1$$

E
$$4 : 1$$

Solution

$$\text{For the first car}:$$

Initial velocity = $$u$$
Final velocity = $$0$$
Deceleration = $$a$$
Using the equation of motion: $$v^2 = u^2 - 2aS$$
$$0=u-2a{ S }_{ 1 }$$
$$\Rightarrow { S }_{ 1 }=\dfrac { u }{ 2a } $$

$$\text{For the second car}:$$
Initial velocity = $$2u$$
Final velocity = $$0$$
Deceleration = $$a$$
Using the equation of motion: $$v^2 = u^2 - 2aS$$
$$0={ \left( 2u \right) }^{ 2 }-2a{ S }_{ 2 }$$
$$\Rightarrow { S }_{ 2 }=\dfrac { 2u }{ a } $$

$$\therefore \dfrac { { S }_{ 1 } }{ { S }_{ 2 } } =\dfrac { \dfrac { u }{ 2a } }{ \dfrac { 2u }{ a } } $$
$$\therefore \dfrac { { S }_{ 1 } }{ { S }_{ 2 } } =\dfrac { 1 }{ 4 } = 1:4$$
Question 7
1 / -0

A body of mass $$2kg$$ has an initial velocity of $$3m/s$$ along $$OE$$ and it is subjected to a force of $$4N$$ along of direction perpendicular to $$OE$$ as shown in figure. The distance travelled by the body in $$4s$$ from $$O$$ will be

A
$$12m$$

B
$$20m$$

C
$$28m$$

D
$$48m$$

Solution

Let $$OE$$ and $$OY$$ be along $$X$$ and $$Y$$ axis respectively,
then $${ u }_{ x }=3m/s,{ a }_{ x }=0\quad $$
and $${ u }_{ y }=0,{ a }_{ y }=\cfrac { 4 }{ 2 } =2m/{ s }^{ 2 }$$
at $$t=4s$$, let $${S}_{x}$$ and $${S}_{y}$$ be the displacement along x-axis and y-axis respectively, then
$${ S }_{ x }={ u }_{ x }=\cfrac { 1 }{ 2 } { a }_{ x }{ t }^{ 2 }=3\times 4+\cfrac { 1 }{ 2 } (0){ (4) }^{ 2 }=12m$$
and
$${ S }_{ y }={ u }_{ y }t+\cfrac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }=0\times 4+\cfrac { 1 }{ 2 } (2){ (4) }^{ 2 }=16m$$
so, the resultant displacement $$S$$ is given as
$$S=\sqrt { { { S }_{ x } }^{ 2 }+{ { S }_{ y } }^{ 2 } } =20m$$
Question 8
1 / -0

A car starts moving along a line, first with an acceleration $$a = 5\ ms^{-2}$$ starting from rest, then uniformly and finally decelerating at the same rate, comes to rest in the total time of $$25\ seconds (t_{1})$$, then average velocity during the time is equal to $$v = 72\ kmph$$. How long does the particle move uniformly?

A
$$5\ seconds$$

B
$$2.5\ hours$$

C
$$1.5\ hours$$

D
$$15\ seconds$$

Solution

Let the initial velocity$$u=0$$

The time taken and the distance traveled during accelerated motion and retarding motion will be same as the acceleration is same

For accelerated motion

$$v=u+at=5t$$

And for retardation

$$0=v-5t$$

$$v=5t$$

As the velocity v will become u for retardation

Total distance traveled

$$=$$ area of $$v-t$$

$$=$$ area$$I+$$ area $$II+$$ area $$III$$

$$2\left( \cfrac { 1 }{ 2 } 5{ t }^{ 2 } \right) +5tx=$$distance traveled during motion

$$x=$$time during which particle remains in uniform motion

$$5{ t }^{ 2 }+5tx=$$speed$$x$$ time

$$5{ t }^{ 2 }+5tx=20\times 25$$

$${ t }^{ 2 }+tx=100\rightarrow 1$$

Also the total time is given $$25$$seconds

Thus $$2t+x=25$$

Thus $$x=25-2t\rightarrow 2$$

Substituting $$2$$ in $$1$$

$${ t }^{ 2 }+t(25-2t)=100$$

$${ t }^{ 2 }-25t+100=0$$

$$t=5$$ or $$t=20$$

Thus the time will be $$t=5$$sec as $$t=20$$s is not possible as it will exceed the given time
The particle remains in motion for $$5$$ seconds
Question 9
1 / -0

In the following displacement (x) vs time (t) graph, at which among the points P, Q and R is the object's speed increasing?

A
P only

B
R only

C
Q and R only

D
P, Q, R

Solution

The slope of the displacement-time graph gives the instantaneous velocity of the object. The magnitude of velocity is speed.
From the graph, at point Q, the slope of the graph is decreasing and thus the speed of the object is also decreasing.
At point R, the slope of the graph is constant and thus the speed of the object is constant.
At point P, the slope of the graph is increasing so the speed of the object increasing.
Question 10
1 / -0

A bullet moving with a speed of $$150$$ $$ms^{-1}$$, strikes a wooden plank. After passing through the plank, its speed becomes $$125$$ $$ms^{-1}$$. Another bullet of the same mass and size strikes the plank with a speed of $$90$$ $$ms^{-1}$$. Its speed after passing through the plank would be.

A
$$25$$ $$ms^{-1}$$

B
$$35$$ $$ms^{-1}$$

C
$$50$$ $$ms^{-1}$$

D
$$70$$ $$ms^{-1}$$

Solution

Let the magnitude of retardation of the bullet during striking be $$a$$.
Here, since retardation is taking place, we take $$-a$$ instead of $$a$$ in the equation

Using, the third equation of motion,
$$v^2=u^2+ 2as$$,

For the first bullet, we have $$ u=150 \ m/s , v = 125 \ m/s$$

$$\Rightarrow 125^2=150^2-2as$$
$$\Rightarrow 2as= 6875$$ ( Where $$s$$ is size of plank , same for both)

For the second bullet, given $$u = 90 \ m/s$$
Let speed $$v$$ after passing through the plank.

$$\Rightarrow v^2=90^2-2as$$
$$\Rightarrow v^2= 8100-6875=1225$$
$$\Rightarrow v=35 ms^{-1}$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 62

Result

Motion Test - 62

Score

-

Rank

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SHARING IS CARING

Question 1

speed

hiking

needle

direction

Solution

Question 2

$$75\ km$$

$$50\ km$$

$$15\ km$$

$$10\ km$$

Solution

Question 3

10 m/s

5 m/s

20 m/s

15 m/s

Solution

Question 4

6 m

2/9 m

2/3 m

18 m

Solution

Question 5

$$4.9m/s$$

$$9.8m/s$$

$$19.6m/s$$

$$25.1m/s$$

Solution

Question 6

$$1 : 1$$

$$1 : 2$$

$$1 : 4$$

$$2 : 1$$

$$4 : 1$$

Solution

Question 7

$$12m$$

$$20m$$

$$28m$$

$$48m$$

Solution

Question 8

$$5\ seconds$$

$$2.5\ hours$$

$$1.5\ hours$$

$$15\ seconds$$

Solution

Question 9

P only

R only

Q and R only

P, Q, R

Solution

Question 10

$$25$$ $$ms^{-1}$$

$$35$$ $$ms^{-1}$$

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

$$70$$ $$ms^{-1}$$

Solution

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