Motion in A Straight Line Test

Result

Motion in A Straight Line Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

A body is moving with velocity $$1$$ m/s at a height $$3$$ m from the ground. What will be its speed at height $$2$$ m from the ground?

A
4.54 m/s

B
1 m/s

C
6 m/s

D
5.32 m/s

Solution
Question 2
1 / -0

A man of mass 40 kg is standing on a uniform plank of mass 60 kg lying on horizontal frictionless ice. The man walks from one end to the other end of the plank . the distance walked by the man relative to ice is (given length of plank=5m)

A
2 m

B
3 m

C
5 m

D
4 m

Solution

d = $$\dfrac{mL}{m_1 + m_2} $$

$$\dfrac{40\times 5m}{100}$$
= 2m
Question 3
1 / -0

A stone is thrown vertically up from a bridge with velocity $$ 3 ms^{-1} $$ if it strikes the water under the bridge after 2 s, the bridge is at a height of $$ ( g= 10m/s^2) $$

A
$$26 m$$

B
$$16 m$$

C
$$14 m$$

D
$$20 m$$

Solution
Question 4
1 / -0

The graph predicts the condition of

A
Body is undergoing positive acceleration

B
Body is undergoing negative acceleration

C
Uniform velocity

D
Uniform speed

Solution
Question 5
1 / -0

A car starts from rest and has an acceleration $$a = 1 \mathrm { m } / \mathrm { s } ^ { 2 }$$.A truck is moving with a uniform velocity of $${ 6 } \mathrm { m } / \mathrm { s }$$. At what distance will the car overtake the truck? direction (at $$t = 0$$ both start their motion in the same direction from the same position)

A
36 m

B
8 m

C
32 m

D
4 m

Solution

At the $$moment$$ when the car will be $$overtaking $$ the truck their $$velocitis$$ will be $$same.$$
so suppose after $$t$$ second from the start of the motion the car overtake the truck then the velocity of the car at that moment will be $$v=0+at=t\times 1m/s^2=t$$ $$m/s$$
and the velocity of the truck at that moment will be $$u=6m/s (constant)$$
both should be same i.e $$u=v$$ or $$6=t$$ or $$t=6$$ $$second$$
so the distance covered by truck $$\text{oe car because both will be at same place ast that point of time }$$ in this time
will be $$s=vt=6\times 6=36meter$$
Question 6
1 / -0

A body is started from rest with acceleration $$2m/{s}^{2}$$ till it attains the maximum velocity then retards to rest with $$3m/{s}^{2}$$. If total time taken is $$10$$ second then maximum speed attained is

A
$$12m/s$$

B
$$8m/s$$

C
$$6m/s$$

D
$$4m/s$$

Solution
Question 7
1 / -0

A body of mass 5kg starts from the origin with an initial velocity $$ \overrightarrow u = 30 \hat i + 40 \hat j ms^{-1} $$ if a constant force $$ \overrightarrow F= - ( \hat i + 5 \hat j) N $$ acts on the body , the time in which the y-component of the velocity becomes zero is:

A
$$5 seconds$$

B
$$20 seconds$$

C
$$40 seconds$$

D
$$80 seconds$$

Solution

correct option $$(C)$$
Question 8
1 / -0

A particle is dropped from a tower. It is found that it travels $$45 m$$ in the last second of its journey. Then height of tower is $$ \left(g=10 \mathrm{m} / \mathrm{s}^{2}\right) $$ ?

A
$$125 m$$

B
$$180 m$$

C
$$100 m$$

D
$$55 m$$

Solution

Given,

$$s=45\,m$$

$$a=g$$

Distance travelled in last second

$$S = u + a(n - \dfrac 12)$$

$$45 = 0 + 10(n - \dfrac 12)$$

$$n = 4.5 + 0.5$$

$$n = 5th\, second$$

Total time of flight is $$5$$ seconds

Height of the tower

$$H = 0.5gt^2$$

$$= 0.5 \times 10 \times 5^2=125\,m$$

Hence, height of the tower is $$125\, m$$.
Question 9
1 / -0

Three particles start from origin at the same time with a velocity $$2 ms^{-1}$$ along positive x-axis the second with a velocity $$6 ms^{-1}$$ aling negative y-axis. Find the velocity of the third particle along x=y line so that the three particles may always lie in a straight line

A
$$-3 \sqrt 3$$

B
$$3 \sqrt 2$$

C
$$-3 \sqrt 2$$

D
$$2 \sqrt 2$$
Question 10
1 / -0

Two trains A and B of length 400 m each are moving on two parallel tracks with uniform speed of
20 m/s in the same direction. with A ahead of B, driver of B accelerates at $$1m/{ s }^{ 2 }$$ in order to overtrake. if after 50 sec, the guard of B just brushes past driver of A, what was original distance between drivers of the trains?

A
$$450 m$$

B
$$1250 m$$

C
$$850 m$$

D
$$1650 m$$

Solution

The relative speed of $$B$$ wrt $$A$$ will be $$V_{BA}=V_B-V_A=20-20=0$$
their relative acceleration will be $$a_{BA}=a_B-a_A=1-0=1m/s^2$$
the distance traveled will be in $$s=ut+\dfrac{1}{2}at^2=0+\dfrac{1}{2}1\times (50)^2=1250meter$$
if initial separation was $$x$$ then $$x+400m=s$$ or $$x=1250-400=850m$$
because now the drivers are side by side of each other so the length of train A was added.