Motion in A Straight Line Test

Result

Motion in A Straight Line Test - 19

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 average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity. The velocity of the body as it reaches the ground is numerically equal to:

A
$$g$$

B
$$\dfrac{g}{2}$$

C
$$\dfrac{g}{\sqrt{2}}$$

D
$$\sqrt{2}g$$

Solution

We have average velocity $$= \dfrac{g}{2}$$
Since it is a free fall, average velocity, $$v=\dfrac{S}{t}= \dfrac{\dfrac{1}{2}gt^2}{t} = \dfrac{g}{2}$$, where $$t$$ is the time of descent and initial velocity is zero.

We know that for a free fall, velocity with which it reaches the ground: $$V = gt$$, where $$t$$ is time of descent .

Substituting the same in the above equation for average velocity:
$$\dfrac{V}{2} = \dfrac{g}{2} $$
$$\therefore V=g$$
Question 2
1 / -0

A body leaving a certain point $$O$$ moves with an acceleration which is constant. At the end of the first second after it left $$O$$, its velocity is $$1.5m/\sec$$. At the end of the sixth second after it left $$O$$ the body stops momentarily and begins to move backwards. The distance traveled by the body before it stops momentarily is

A
$$9 m$$

B
$$18 m$$

C
$$54 m$$

D
$$36 m$$

Solution

$$v=u+at$$
$$1.5=u+5a...........(i)$$
$$0=u+6a.............(ii)$$
From (i) and (ii)
$$0=1.5+a(6-5)$$
$$a=-1.5m/s^2$$
Om substituting $$a=-1.5m/s^2$$ in equation (ii)
$$0=u+6(-1.5)$$
$$0=u-9$$
$$u=9m/s$$
The distance traveled by the body in its forward journey is $$s_1$$ can be calculated as follows,
$$s_1=ut+\frac{1}{2}at^2$$
$$s_1=9\times6+\frac{1}{2}\times(-1.5)\times6^2$$
$$s_1=54+(-1.5\times18)$$
$$s_1=27m$$
The distance traveled by the body in its forward journey $$(s_1)$$= the distance traveled by the body in its backward journey$$(s_2)=27m$$
So, the total distance traveled by the body in forward journey and its backward journey$$=s_1+s_2=27+27=54m$$
Question 3
1 / -0

A driver takes $$0.20s$$ to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of $$54 km/h$$ and the brakes causes a deceleration of $$ 6.0 \ { m }/{ { s }^{ 2 } }$$, find the distance traveled by the car after he sees the need to put the brakes on.

A
$$18.63 m$$

B
$$20 m$$

C
$$26.85 m$$

D
$$27.67 m$$
Question 4
1 / -0

A train accelerates from rest at a constant rate $$a_{1}$$ for distance $$S_{1}$$ and time $$t_{1}$$. After that it decelerates to rest at a constant rate $$a_{2}$$ for distance $$S_{2}$$ at time $$t_{2}$$. Then the correct relation among the following is

A
$$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{1}}{a_{2}} = \frac{t_{1}}{t_{2}}$$

B
$$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{2}}{a_{1}} = \frac{t_{1}}{t_{2}}$$

C
$$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}}$$

D
$$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{2}}{a_{1}} = \frac{t_{2}}{t_{1}}$$

Solution

Given that the particle was at rest at $$t\ =\ 0$$
$$s_1 = u_1t + \frac {1}{2} a_1t_1^2 .... (1)$$
$$\Rightarrow v_1 = u_1 + a_1t_1 .... (2)$$
$$v_1 \rightarrow $$ final speed after $$t_1$$

Now, final speed after this is $$ v_2$$
And $$v_1 = v_2$$
$$\Rightarrow v_2 = u_2 - a_2t_2$$
$$\Rightarrow v_1 = a_2t_2 .... (3)$$
$$\Rightarrow v_2^2 = u_2^2 - 2a_2s_2$$
$$\Rightarrow v_1^2 = 2a_2s_2$$

From (2), $$a_1^2t_1^2 = 2a_2s_2 $$

$$\Rightarrow s_2 = \dfrac {a_1^2t_1^2}{2a_2}$$

$$\Rightarrow s_2 = \dfrac {a_1 (2s_a)}{2\frac {a}{2}a_2}$$

$$\Rightarrow \dfrac {eq^2 (2)}{eq^n (3)} = \dfrac {a_1t_1}{a_2t_2} = 1$$

$$\Rightarrow \dfrac {a_1}{a_2} = \dfrac {t_2}{t_1}$$

$$\Rightarrow \dfrac {a_1}{a_2} = \dfrac {t_2}{t_1} = \dfrac {s_2}{s_1}$$
Question 5
1 / -0

A proton in a uniform electric field moves along a straight line with constant acceleration starting from rest. If it attains a velocity $$4\times 10^3$$ $$km/s$$ at a distance of $$2\;cm$$, the time required to reach the given velocity is :

A
$$10^{-3}\;s$$

B
$$10^{-6}\;s$$

C
$$10^{-8}\;s$$

D
$$10^{-5}\;s$$

Solution

$$v^2 = u^2 + 2as$$
$$u = 0, s = 2 \times 10^{-2} mt$$
$$a = \dfrac {v^2}{2s} = \dfrac {(4 \times 10^3 \times 10^3)^2}{2 \times 2 \times 10^{-2}} m/s^2$$
$$a = 4 \times 10^{14} $$

Now, $$ v = u + at$$
$$\dfrac {v}{a} = t$$
$$t = \dfrac{4 \times 10^6}{4 \times 10^{14}} = 10^{-8}s$$
Question 6
1 / -0

From a building two balls A & B are thrown such that A is thrown upwards and B downwards (both vertically with same speed). If $$V_{A}$$ and $$V _{B}$$ are their respective velocities on reaching the ground then :

A
$$V_{B} < V_{A}$$

B
$$V_{A} = V_{B}$$

C
$$V_{A} < V_{B}$$

D
Their velocities depends on their masses

Solution

For A:
$$s_A = vt_1 - \dfrac {1}{2} gt_1^2$$
$$s_B = -vt_2 - \dfrac {1}{2} gt_2^2$$

We have, $$s_A = s_B$$

So, on solving we get:
$$t_2 - t_1 = \dfrac {2v}{g}$$ ...........(1)
Also,
$$v_A = u_A - gt_1$$
$$v_B = u_B - gt_2$$

Using $$(1)$$, we get:

$$ \dfrac {v_A}{v_B} = \dfrac {v - gt_1}{-v -gt_2}$$$$= \dfrac {v - gt_1}{-v -g (t_1 + \dfrac {2v}{g})} = 1$$

$$\Rightarrow v_A = v_B$$
Question 7
1 / -0

It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him 3 minutes to walk up. How long will it take for the passenger to arrive at the top if he walks up the moving escalator?

A
$$30\ sec$$

B
$$45\ sec$$

C
$$40\ sec$$

D
$$35\ sec$$

Solution

Let the distance to the top be d.
therefore the speed of the man with escalator not moving $$=(\dfrac{d}{1}) m/min$$
therefore the speed of the escalator with the man at rest$$=(\dfrac{d}{3})$$m/min
when both the main and the escalator are moving, effective speed$$= (\dfrac{d}{3} +d)$$ m/min
$$= \dfrac{4d}{3}$$ m/min
Therefore time taken when both men and escalator are moving$$= \dfrac{distance}{speed} = \dfrac{d}{\dfrac{4d}{3}}$$
$$\dfrac{3}{4} min = 45$$seconds
Hence (B) is correct answer
Question 8
1 / -0

Two cars are travelling towards each other on a straight road at velocities $$15\ m/s$$ and $$16\ m/s$$ respectively. When they are $$150\ m$$ apart, both the drivers apply the brakes and the cars decelerate at $$3\ m/s^{2}$$ and $$4\ m/s^{2}$$ until they stop. Separation between the cars when they come to rest is :

A
$$86.5 m$$

B
$$89.5 m$$

C
$$85.5 m$$

D
$$80.5 m$$

Solution

We know, $$v^2=u^2-2as$$

Given,

Initial velocity of car 1, $$u_1=15\ m/s$$, Acceleration, $$a=3\ m/s^2$$

Initial velocity of car 2, $$u_2=16\ m/s$$, Acceleration, $$a=4\ m/s^2$$

For car 1,

Final Velocity, $$v_1 = 0\ m/s$$

$$-15^2 = -2(3) \times s$$

$$\therefore s_1= 37.5$$ m

For car 2,

Final Velocity, $$v_2 = 0\ m/s$$

$$-16^2 = -2(4) \times s$$

$$\therefore s_2= 32$$ m

Total displacement,

$$s_1+s_2 = 37.5+32 = 69.5$$ m

$$\therefore$$ separation between them,

$$= 150-69.5 =80.5$$ m
Question 9
1 / -0

From an elevated point P, a stone is projected vertically upwards. When it reaches a distance $$d$$ below P, its velocity is doubled. The greatest height reached by it above P is :

A
$$d/3$$

B
$$3d$$

C
$$2d$$

D
$$d/2$$

Solution

$$v -u =gt$$
$$2u-u =gt$$

$$\Rightarrow t = \dfrac{u}{g}$$ where, t is the time to go from u (downward) to 2u (downward).

$$t_{total} = \dfrac{2u}{g} + \dfrac{u}{g}= \dfrac{3u}{g}$$ since $$\dfrac{u}{g}$$ is time to go up and same time to come down.

$$(2u)^2 - u^2 = 2g d$$

$$\Rightarrow d = \dfrac{3u^2}{2g}$$
Maximum height, $$H = \dfrac {u^2}{2g} = \dfrac {d}{3}$$
Question 10
1 / -0

A bus starts from rest and moves with a uniform acceleration of $$1\;ms^{-2}$$. A boy $$10\;m$$ behind the bus at the start runs at a constant speed and catches the bus in $$10\;s$$. Speed of the boy is :

A
$$10 ms^{-1}$$

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

C
$$ 6 ms^{-1}$$

D
$$4 ms^{-1}$$

Solution

Solution 2
Distance travelled by bus in $$10s \Rightarrow \dfrac {1}{2} \times 1 \times 10^2 = 50 m$$
D $$=$$ Distance boy has to cover $$= 10 + 50 = 60 m$$
Speed of boy $$=\dfrac {D}{T} = \dfrac {60}{10} = 6\ m/s$$

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 - 19

Result

Motion in A Straight Line Test - 19

Score

-

Rank

-

SHARING IS CARING

Question 1

$$g$$

$$\dfrac{g}{2}$$

$$\dfrac{g}{\sqrt{2}}$$

$$\sqrt{2}g$$

Solution

Question 2

$$9 m$$

$$18 m$$

$$54 m$$

$$36 m$$

Solution

Question 3

$$18.63 m$$

$$20 m$$

$$26.85 m$$

$$27.67 m$$

Question 4

$$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{1}}{a_{2}} = \frac{t_{1}}{t_{2}}$$

$$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{2}}{a_{1}} = \frac{t_{1}}{t_{2}}$$

$$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}}$$

$$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{2}}{a_{1}} = \frac{t_{2}}{t_{1}}$$

Solution

Question 5

$$10^{-3}\;s$$

$$10^{-6}\;s$$

$$10^{-8}\;s$$

$$10^{-5}\;s$$

Solution

Question 6

$$V_{B} < V_{A}$$

$$V_{A} = V_{B}$$

$$V_{A} < V_{B}$$

Their velocities depends on their masses

Solution

Question 7

$$30\ sec$$

$$45\ sec$$

$$40\ sec$$

$$35\ sec$$

Solution

Question 8

$$86.5 m$$

$$89.5 m$$

$$85.5 m$$

$$80.5 m$$

Solution

Question 9

$$d/3$$

$$3d$$

$$2d$$

$$d/2$$

Solution

Question 10

$$10 ms^{-1}$$

$$ 1 ms^{-1}$$

$$ 6 ms^{-1}$$

$$4 ms^{-1}$$

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.