Motion in A Straight Line Test

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Motion in A Straight Line Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

A stone is dropped into the water from a bridge $$44.1 m$$ above the water. another stone is thrown vertically downward $$1 $$second later. both strike the water simultaneously . then initial speed of the second stone is

A
$$ 24.5 ms^{-1} $$

B
$$ 4.9 ms^{-1} $$

C
$$ 9.8 ms^{-1} $$

D
$$ 12.25 ms^{-1} $$

Solution
Question 2
1 / -0

A bullet moving at 20 m/sec. it strike a wooden plank and pentrates 4 cm before coming to stop. the time taken to stop is

A
0.08 sec

B
0.16 sec

C
0.004 sec

D
0.02 sec

Solution

Given,
$$u=20m/s$$
$$s=4cm=0.04m$$
$$v=0m/s$$
From 3rd equation of motion,
$$2as=v^2-u^2$$
$$2\times a\times 0.04=0^2-20\times 20$$
$$a=-\dfrac{20\times 20}{2\times 0.04}$$
$$a=-5000m/s^2$$
From 1st equation of motion,
$$v=u+at$$
$$0=20-5000t$$
$$t=\dfrac{20}{5000}=0.004sec$$
Question 3
1 / -0

A particle is dropped from a tower. It is found that it travels 55m in the last second of its journey. Then height of the tower is (g= 10 m/s2) ?

A
$$125 m$$

B
$$180 m$$

C
$$100 m$$

D
$$55 m$$

Solution
Question 4
1 / -0

A ball is dropped from a roof to the ground 8.0 m below. A rock is thrown down from the roof 0.6 s later. If they both hit the ground at the same time, what was the intitial speed of the rock?

A
4.43 m/s

B
8.43 m/s

C
16.43 m/s

D
12.43 m/s

Solution
Question 5
1 / -0

If initial velocity of an object is 'u' and acceleration is 'a' then find the distance travelled in nth second.

A
$$S_n = un + \dfrac {an^2} {2}$$

B
$$S_n = un + an^{2}$$

C
$$S_n = u + \dfrac {a} {2} (2n - 1)$$

D
$$S_n = (u + \dfrac a 2)n^2$$

Solution
Question 6
1 / -0

A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is $$S_1$$ and that covered in the first 20 seconds is $$S_2$$ , then

A
$$S_2= 3S_1$$

B
$$S_2= 4S_1$$

C
$$S_2= S_1$$

D
$$S_2= 2S_1$$

Solution
Question 7
1 / -0

A body dropped from the top of a tower covers $$7/16$$ of the total height in the last second of its fall. The time of fall is

A
$$2$$ sec

B
$$4$$ sec

C
$$1$$ sec

D
$$(\cfrac{50}{7})$$ sec

Solution
Question 8
1 / -0

In this problem 'up' is taken to be positive. A ball thrown vertically upward with an initial velocity of 19.6 $$ \mathrm{m} / \mathrm{s} $$, after 5 s has a velocity of

A
$$

19.6 \mathrm{m} / \mathrm{s}

$$

B
-$$

19.6 \mathrm{m} / \mathrm{s}

$$

C
$$

29.4 \mathrm{m} / \mathrm{s}

$$

D
-$$

29.4 \mathrm{m} / \mathrm{s}

$$

Solution

Here,

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

$$t=5\,s$$

$$a=-g=-9.8\,m/s^2$$

$$v=?$$

Using $$v=u+at$$

We gaet,

$$v=19.6-(9.8\times 5)=-29.4m/s$$
Question 9
1 / -0

What is the angle between instantaneous displacement and acceleration during the retarded motion

A
$$Zero$$

B
$$
\pi
$$

C
$$
\frac{\pi}{2}
$$

D
$$
\frac{\pi}{4}
$$

Solution
Question 10
1 / -0

A person travelled a distance of 3 km along a straight line in the North direction .Then he travelled 2 km in west direction and then 5 km in south direction.The magnitude of the displace-ment of this person would be ___________.

A
$$2\sqrt { 2 } $$ Km

B
$$3\sqrt { 2 } $$ Km

C
$$4\sqrt { 2 } $$ Km

D
10 Km

Solution

IA person walked 3 km in north, 2km west and 5 km in South. Then, the map becomes like:
Then now we have .
P=5–3=2km b=2km
So H^2=P^2+B^2,
H^2= 4+4
H^2=8
Or H=√8
The displacement travelled by person is 2√2 km or 2.82km