Motion Test

Result

Motion Test - 66

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

A truck traveling due north at $$50 \ km/hr$$ turns west and travels at the same speed. What is the change in magnitude of speed?

A
$$0$$

B
$$100$$

C
$$5$$

D
$$50$$

Solution

When the direction is changed then velocity changes but the speed or magnitude of velocity remains constant.
In this case, the truck changes only the direction of motion, not the magnitude.
So change in the magnitude of velocity is zero.
Question 2
1 / -0

A body starts from rest under uniform acceleration. If distance covered by it in $$3^{rd}$$ second is $$3m$$, then the distance covered in $$5^{th}$$ second will be

A
$$5m$$

B
$$3m$$

C
$$\cfrac{27}{5}m$$

D
$$\cfrac{24}{5}m$$

Solution

$$ s = u + \cfrac{a}{2} (2n -1)$$
As, $$ 3 = 0 + \cfrac{a}{2} [2 \times 3 -1]$$
$$ \therefore a = \cfrac{6}{5}$$
Now,$$ s' = 0 + \cfrac{6}{2\times 5} (2\times 5 -1)$$
$$ = \cfrac{6}{2 \times 5} \times 9 = \cfrac{54}{10}$$
$$ = \cfrac{27}{5}m$$
Question 3
1 / -0

A body is dropped from a height $$39.2\ m$$. After it crosses the half distance, the acceleration due to gravity ceases to act. Then the body will hit the ground with a velocity of (Take $$g=9.8\ ms^{-2}$$)

A
$$19.6\ ms^{-1}$$

B
$$20\ ms^{-1}$$

C
$$1.96\ ms^{-1}$$

D
$$196\ ms^{-1}$$

Solution

Given that,

Height $$h = 39.2\ m$$

Initial velocity $$u = 0$$

Now, from equation of motion

After crosses half distance, the acceleration due to gravity ceases to act

So,

$$ h=\dfrac{39.2}{2} $$

$$ h=19.6\,m $$

$$ {{v}^{2}}-{{u}^{2}}=2gh $$

$$ {{v}^{2}}=2\times 9.8\times 19.6 $$

$$ v=19.6\,m/s $$

Hence, the velocity is $$19.6\ m/s$$
Question 4
1 / -0

A body travels a distance x in first two second and distance y in next two second. The relation between $$x$$ and $$y$$ is

A
$$y=4x$$

B
$$y=x$$

C
$$y=3x$$

D
$$y=2x$$

Solution

$$\rightarrow $$ Distance Travelled in $$2\ sec$$
$$S= ut + \dfrac{1}{2} at^{2} $$u=0$$ $$t=2$$
$$\therefore x=0+ \dfrac{1}{2} a2^{2} = x= 2a \Rightarrow a = \dfrac{x}{2}$$
Distance travelled in $$4\ sec$$ Distance travelled in $$2\ sec =$$ Distance travelled in next $$2\ sec$$
$$\therefore y =\dfrac{a}{2} (4)^{2}- \dfrac{a}{2} (2)^{2}= 8a-2 a = 6a$$
$$\therefore y= 6a \Rightarrow y=6 \dfrac{(x)}{2}$$
$$\left[ \because x=2a \ a=\dfrac{x}{2} \right]$$
$$\therefore y= 3x$$
Answer is $$C$$
Question 5
1 / -0

The speed-time graph of a particle moving along a solid cover is shown below. The distance traversed by the particle from $$t = 0$$ to $$t = 3$$ is:-

A
$$\dfrac{9}{2} m$$

B
$$\dfrac{9}{4} m$$

C
$$\dfrac{10}{3} m$$

D
$$\dfrac{10}{5} m$$

Solution

Distance covered is equal to the area under the graph in given interval of time.

$$\Rightarrow Distance = area = \dfrac{1}{2} \times 3\times 1.5 = \dfrac{9}{4}m$$

Thereefore, B is correct option.
Question 6
1 / -0

Ball $$A$$ is thrown up vertically with speed $$10 \ m/s$$. At the same instant another ball $$B$$ is released from rest at height $$h$$. At time $$t$$, the speed of $$A$$ relative to $$B$$ is

A
$$10 \ m/s$$

B
$$10 - 2 \ gt$$

C
$$\sqrt{10^2 - 2gh}$$

D
$$10 - gt$$

Solution

$$u_A=10 ,\ u_B=0\ ,\ a_A=-g\ , a_B=-g $$
$$u_{AB}=10\ , \ a_{AB}=0 $$
$$v_{AB}=u_{AB}+a_{AB}t$$
$$v_{AB}=10m/s$$
Question 7
1 / -0

A car covers a distance of $$2 \ km$$ in $$2.5$$ minutes. If it covers half of the distance with speed $$40 \ km/hr$$, the rest distance it shall cover with a speed of:

A
$$56 \ km/hr$$

B
$$60 \ km/hr$$

C
$$48 \ km/hr$$

D
$$50 \ km/hr$$

Solution

Given that,

Time $$t=2.5\min $$

Distance $$d=2km$$

Now, firstly for half distance

$$ v=\dfrac{d}{t} $$

$$ t=\dfrac{d}{v} $$

$$ t=\dfrac{60\times 60}{40} $$

$$ t=90\,s $$

Now, total time is

$$ t=150-90 $$

$$ t=60\,s $$

Now, the speed is

$$ v=\dfrac{d}{t} $$

$$ v=\dfrac{1\times 60}{1} $$

$$ v=60\,km/h $$

Hence, the speed is $$60 km/h$$
Question 8
1 / -0

In uniform circular motion the

A
acceleration is parallel (or antiparallel) to the velocity

B
acceleration is perpendicular to the velocity

C
acceleration is vertical, while the velocity can be any direction

D
acceleration is vertical and the velocity is horizontal

Solution
Question 9
1 / -0

A particle travels half of the distance of a straight journey with a speed $$6 \ m/s$$. The remaining part of the distance is covered with speed $$2 \ m/s$$ for half of the time and with speed $$4 m/s$$ for other half of time. The average speed of the particle is :

A
$$3 \ m/s$$

B
$$4 \ m/s$$

C
$$\dfrac{3}{4} \ m/s$$

D
$$5 \ m/s$$

Solution

Let total distance covered in the entire journey is $$2x$$.
Let $$t_1$$ is the time to cover half of the distance with a speed of $$6 \ m/s$$.

So,$${{t}_{1}}=\dfrac{x}{6}............(1)$$

Let, for the second part total time is $$t_2$$.
Let the distance covered in first half time with a speed of $$2 \ m/s$$ is $$x_1$$. So, $${{x}_{1}}=\dfrac{{{t}_{2}}}{2}\times 2={{t}_{2}}............(2)$$

For the second half the distance covered with a speed of $$4 \ m/s$$is $$x_2$$. So, $${{x}_{2}}=\dfrac{{{t}_{2}}}{2}\times 4=2{{t}_{2}}..........(3)$$

Now,

$$ {{x}_{1}}+{{x}_{2}}=x $$

$$ {{t}_{2}}+2{{t}_{2}}=x $$

$$ {{t}_{2}}=\dfrac{x}{3}............(3) $$
Hence, $$total \ time = t_1+t_2 = \dfrac{x}{6} + \dfrac{x}{3}$$

Now , $$ average\,\,speed=\dfrac{total\,\,distance}{total\,\,time} $$

$$ s=\dfrac{2x}{\dfrac{x}{6}+\dfrac{x}{3}}=\dfrac{2x}{3x}\times 6 $$

$$ s=4\,m/s $$
Question 10
1 / -0

In a uniform circular motion -

A
both velocity and acceleration are constant.

B
acceleration and speed are constant but velocity changes.

C
both acceleration and velocity change

D
both acceleration and speed are constant

Solution

For a body moving in uniform circular motion the speed attained by the object will remain constant not the velocity as velocity depends on the direction of motion, and in circular motion the direction of the object changes at every point. As speed remains constant the acceleration will remain constant.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Motion Test - 66

Result

Motion Test - 66

Score

-

Rank

-

SHARING IS CARING

Question 1

$$0$$

$$100$$

$$5$$

$$50$$

Solution

Question 2

$$5m$$

$$3m$$

$$\cfrac{27}{5}m$$

$$\cfrac{24}{5}m$$

Solution

Question 3

$$19.6\ ms^{-1}$$

$$20\ ms^{-1}$$

$$1.96\ ms^{-1}$$

$$196\ ms^{-1}$$

Solution

Question 4

$$y=4x$$

$$y=x$$

$$y=3x$$

$$y=2x$$

Solution

Question 5

$$\dfrac{9}{2} m$$

$$\dfrac{9}{4} m$$

$$\dfrac{10}{3} m$$

$$\dfrac{10}{5} m$$

Solution

Question 6

$$10 \ m/s$$

$$10 - 2 \ gt$$

$$\sqrt{10^2 - 2gh}$$

$$10 - gt$$

Solution

Question 7

$$56 \ km/hr$$

$$60 \ km/hr$$

$$48 \ km/hr$$

$$50 \ km/hr$$

Solution

Question 8

acceleration is parallel (or antiparallel) to the velocity

acceleration is perpendicular to the velocity

acceleration is vertical, while the velocity can be any direction

acceleration is vertical and the velocity is horizontal

Solution

Question 9

$$3 \ m/s$$

$$4 \ m/s$$

$$\dfrac{3}{4} \ m/s$$

$$5 \ m/s$$

Solution

Question 10

both velocity and acceleration are constant.

acceleration and speed are constant but velocity changes.

both acceleration and velocity change

both acceleration and speed are constant

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.