Motion Test

Result

Motion Test - 61

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

The displacement - time graph of a particle moving along a straight line is given below. Find the time at which its velocity is equal to zero.

A
$$0$$

B
$$2$$

C
$$4$$

D
None of these

Solution

The slope of displacement - time graph gives us the velocity. From the given graph, we see that the slope at $$t=2$$ is $$0$$. Hence, velocity at $$t=2$$ is $$0$$.
Question 2
1 / -0

If $${ v }^{ 2 }={ u }^{ 2 }+2as$$, then $$u =$$

A
$${ v }^{ 2 }-2as$$

B
$$\pm \sqrt { 2as-{ v }^{ 2 } } $$

C
$$\pm \sqrt { { v }^{ 2 }-2as } $$

D
$$\pm \sqrt { { v }^{ 2 }+2as } $$

Solution

Given : $$v^2=u^2+2as$$
$$\implies$$ $$u^2=v^2-2as$$
Taking square root on both the sides we get
$$u=\pm \sqrt{v^2-2as}$$
Hence, option C is correct.
Question 3
1 / -0

A dragster accelerates to a speed of $$112 m/s$$ over a distance of $$398 m$$. Determine the acceleration (assume uniform) of the dragster.

A
$$5.8 m/s^2$$

B
$$11.8 m/s^2$$

C
$$15 m/s^2$$

D
$$15.8 m/s^2$$

Solution

We assume that the dragster accelerates from rest. So, the initial velocity of it is $$u=0 m/s$$
Given, final velocity is $$v=112 m/s$$ and traveled distance, $$d=398 m$$
Let $$a$$ be the acceleration.
Using formula $$v^2-u^2=2ad$$,
$$\therefore $$ $$(112)^2-0^2=2a(398)$$
$$\implies a=15.76 \sim 15.8 m/s^2$$
Question 4
1 / -0

An aircraft has a lift-off of $$120km/h$$. What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of $$240m$$? (in $$m/s^2$$)

A
$$2.2$$

B
$$2.3$$

C
$$2.4$$

D
$$2.5$$

Solution

Here, initial velocity of aircraft is $$u=0 m/s$$.
Final velocity is $$v=120 km/hr=\dfrac{120\times 1000}{3600}=33.33 m/s$$
Traveled distance is $$d=240 m$$
If $$a$$ be the required acceleration.
Using $$v^2-u^2=2ad$$,
$$(33.33)^2-0^2=2a(240)$$
$$\implies$$ $$a=2.3 m/s^2$$
Question 5
1 / -0

Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of $$26.2 m/s$$. Determine the height to which the vase will rise above its initial height.

A
$$34.0 m$$

B
$$39.0 m$$

C
$$35.0 m$$

D
$$37.0 m$$

Solution

The initial height of the vase is the top most point of its motion and at that point velocity reaches to zero. So, the final velocity is $$v=0 m/s$$.
Here, acceleration is $$a=-g=-9.8 m/s^2$$ (minus sign for the motion against gravity)
Initial velocity is $$u=26.2 m/s$$.
If $$h$$ be the required height.
Using formula $$v^2-u^2=2ad$$,
$$0^2-(26.2)^2=2(-9.8)h$$
$$\implies$$ $$h=35.0 m $$
Question 6
1 / -0

Directions For Questions

A missile is launched into the air at an initial velocity of $$80 m/s $$. It is moving with constant velocity until it reaches $$1000m$$, when the engine fails.

...view full instructions

How long does it take it to reach $$1000m$$ $$in \ seconds$$ ?

A
$$12$$

B
$$12.5$$

C
$$11$$

D
$$13$$

Solution

Velocity of missile $$v = 80 \ m/s$$
Distance covered $$d = 1000 \ m$$
Time taken $$t = \dfrac{d}{v} = \dfrac{1000}{80} = 12.5 \ s$$
Question 7
1 / -0

A stone dropped from the top of the tower reaches ground in $$4\ sec$$. Height of the tower is $$(g = 10m/s^{2})$$

A
$$20m$$

B
$$40m$$

C
$$60m$$

D
$$80m$$

Solution

Initial speed of the stone $$u =0 m/s$$

Time taken by the stone to reach the ground $$t = 4 s$$

Height of the tower $$H = ut +\dfrac{1}{2}at^2$$

$$\therefore$$ $$H = 0 +\dfrac{1}{2}\times 10\times 4^2$$

$$\implies$$ $$H = 80 m$$
Question 8
1 / -0

A bullet when fired into a target loses half of its velocity after penetrating 20 cm. Further distance of penetration before it comes to rest is :

A
6.66 cm

B
3.33 cm

C
12.5 cm

D
10 cm

E
5 cm

Solution

Given, Initial velocity, $$U=v$$
Final velocity, $$V=\frac {v}{2}$$
Distance, $$s=20 cm$$

Let, the further distance of penetration before it comes to rest be x.
$$V^2 =U^2 -2as$$
$$\left ( \frac {V}{2} \right )^2=V^2 -2a\times 20$$
$$40 a=V^2-\frac {V^2}{4}$$
$$40 a=\frac {3V^2}{4}$$ ...(i)

and $$V^2 =u^2 +2as$$
$$v^2=0+2a\times (20 +x)$$
$$v^2 = 2\times \frac {3}{160}v^2\times (20+x)$$ [From Eq. (i)]
$$1=\frac {3}{80}(20+x)$$
$$\frac {80}{3}=20 +x$$
$$x=\frac {80}{3}-20$$
$$x=\frac {80-60}{3}$$
$$x=\frac {20}{3}=6.66cm$$
Question 9
1 / -0

The meter that is used to measure the distance moved by the vehicle is known as _____. Fill in the blank.

A
Speedometer

B
Odometer

C
Chronometer

D
Ammeter

Solution

Odometer is an instrument used in the vehicle to measure the distance cover by it.
Question 10
1 / -0

Fill in the blank:
The _______ circular motion describes as the motion of an object in a circular path with a _______ speed.

A
Non-uniform, constant

B
Uniform, constant

C
Non-uniform, varying

D
Uniform, Varying

Solution

The object is said to be in an uniform circular motion if the speed of the object moving in the circular path is constant.
Whereas the object is said to be in a non-uniform circular motion if the speed of the object moving in the circular path is varying (either increasing or decreasing).

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Motion Test - 61

Result

Motion Test - 61

Score

-

Rank

-

SHARING IS CARING

Question 1

$$0$$

$$2$$

$$4$$

None of these

Solution

Question 2

$${ v }^{ 2 }-2as$$

$$\pm \sqrt { 2as-{ v }^{ 2 } } $$

$$\pm \sqrt { { v }^{ 2 }-2as } $$

$$\pm \sqrt { { v }^{ 2 }+2as } $$

Solution

Question 3

$$5.8 m/s^2$$

$$11.8 m/s^2$$

$$15 m/s^2$$

$$15.8 m/s^2$$

Solution

Question 4

$$2.2$$

$$2.3$$

$$2.4$$

$$2.5$$

Solution

Question 5

$$34.0 m$$

$$39.0 m$$

$$35.0 m$$

$$37.0 m$$

Solution

Question 6

$$12$$

$$12.5$$

$$11$$

$$13$$

Solution

Question 7

$$20m$$

$$40m$$

$$60m$$

$$80m$$

Solution

Question 8

6.66 cm

3.33 cm

12.5 cm

10 cm

5 cm

Solution

Question 9

Speedometer

Odometer

Chronometer

Ammeter

Solution

Question 10

Non-uniform, constant

Uniform, constant

Non-uniform, varying

Uniform, Varying

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.