Motion in A Straight Line Test

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Motion in A Straight Line Test - 49

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Question 1
1 / -0

A body of mass $$500 g$$ is thrown upward with a velocity $$20 m/s $$ and reaches back to the surface of a planet after $$20$$ sec Then the weight of the body on that planet is :

A
$$2N$$

B
$$4 N$$

C
$$5 N$$

D
$$1 N$$

Solution

The acceleration when the mass is going up is given as,

$${v^2} = {u^2} + 2as$$

$${\left( {20} \right)^2} = {\left( 0 \right)^2} + 2as$$

$$a = \dfrac{{200}}{s}$$ (1)

The acceleration when the mass is going down is given as,

$$s = \dfrac{1}{2}a{t^2}$$

$$s = \dfrac{1}{2}a{\left( {10} \right)^2}$$

$$a = \dfrac{s}{{50}}$$ (2)

From equation (1) and (2), we get

$$\dfrac{{200}}{s} = \dfrac{s}{{50}}$$

$$s = 100\;{\rm{m}}$$

Substitute the value of $$s$$ in the equation (2), we get

$$a = \dfrac{{100}}{{50}}$$

$$ = 2\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$

The weight is given as,

$$W = 0.5 \times 2$$

$$ = 1\;{\rm{N}}$$

Thus, the weight of the body on that planet is $$1\;{\rm{N}}$$.
Question 2
1 / -0

A particle experiences constant acceleration for $$20$$ seconds after starting from rest. If it travels a distance $$s_1$$ in the first $$10$$ seconds and distance $$s_2$$ in the next $$10$$ second then:

A
$$s_2=s_1$$

B
$$s_2=2s_1$$

C
$$s_2=3s_1$$

D
$$s_2=4s_1$$

Solution

Distance traveled in first $$10 seconds$$

$$S_1 = u t + \dfrac{1}{2} a t^2 = 0 t + \dfrac{1}{2} a 10^2 = 50 a$$

Total distance traveled in 20 seconds =

$$ S = u t + \dfrac{1}{2} a t^2 = 0 t + \dfrac{1}{2} a 2062 = 200 a$$

Distance traveled from $$10s$$ to $$20s$$$$ = S_2 = 200 a - 50 a = 150 a$$
Hence answer is C.
Question 3
1 / -0

Two particles start moving from the same point along the same stright line. The first moves with constant velocity v and the second with constant acceleration a. During the time that elapses before the second catches the first, the greater distance between the particles is

A
$$\dfrac{v^2}{a}$$

B
$$\dfrac{v^2}{2a}$$

C
$$\dfrac{2v^2}{a}$$

D
$$\dfrac{v^2}{3a}$$

Solution

A distance between the two particles is maximum when both have the same velocity
let the second move has a velocity v at time t
$$v=at$$
$$t=\dfrac{v}{a}$$
distance cover by the first paticle $$s_{1}$$ is
= $$s_{1}=vt=v\times \dfrac{v}{a}=\dfrac{v^2}{a}$$
distance cover the particle $$ s_{2}$$
$$=\dfrac{1}{2}\times at^2=\dfrac{v^2}{2a}$$
distance between both the particle is
$$ s_{1}-s_{2}$$
$$= \dfrac{v^2}{a}-\dfrac{v^2}{2a}=\dfrac{v^2}{2a}$$
hence the B option is correct
Question 4
1 / -0

A body lying initially at point (3, 7) starts moving with a constant acceleration of 4i. Its position after 3 s is given by the co-ordinates:

A
(7, 3)

B
(7, 18)

C
(21, 7)

D
(3, 7)

Solution

The initial coordinate of the particle is $$(3,7)$$
x-coordinate is 3 unit
y-coordinate is 7 unit
As the body starts from rest so initial velocity$$=0$$
The acceleration of the body along x-axis$$=4 units$$
The acceleration of the body along y-axis$$=0$$
So, after 3 seconds,
Using equation of motion along x-axis,
$$x=x_0+ut+\frac{1}{2}at^2$$
as $$u=0\quad a=4units\quad t=3seconds$$
$$s=3+\frac{1}{2}\times4\times3^2=21units$$
As there is no motion along y-axis so it will remain unchanged
Hence the final coordinate are $$(21,7)$$
Question 5
1 / -0

If a car at rest accelerates uniformly to a speed of 144 km$$h^{-1}$$ in 20 s, then it covers a distance of:

A
20 m

B
400 m

C
1440 m

D
2880 m

Solution

Given
$$u=0$$
$$v=144kmph=144\times\frac{5}{18}=40 m/s$$
$$t=20sec$$
We know,
$$v=u+at$$
$$40=a\times20$$
$$a=2m/s^2$$
Now,
$$s=ut+\frac{1}{2}at^2$$
$$s=\frac{1}{2}at^2=\frac{1}{2}\times2\times20^2=400m$$
Question 6
1 / -0

A $$50-kg$$ crate is being pushed across a horizontal floor by a horizontal force of $$575\ N$$. If the coefficient of sliding friction is $$0.25$$, what is the acceleration of the crate?

A
Zero

B
$$1\ m/s^{2}$$

C
$$3\ m/s^{2}$$

D
$$6\ m/s^{2}$$

E
$$9\ m/s^{2}$$

Solution

Given,

Force on crate, $$F=\,575\,N$$

Mass of crate, $$m=50\,kg$$

Coefficient of friction, $$\mu =0.25$$

From the equilibrium of forces

$$ma=F-\mu mg$$

$$a=\dfrac{F-\mu mg}{m}$$

$$a=\dfrac{575-0.25\times 50\times 9.81}{50}$$

$$a=9.047\,\,m/{{s}^{2}}$$

Acceleration of block is $$9.047\,\,m/{{s}^{2}}$$
Question 7
1 / -0

An object moving with an speed of 25 m/s is decelerating at a rate given by $$a = -5\sqrt{v}$$, where v is speed at any time. The time taken by the object to come to rest is:

A
1 s

B
5 s

C
2 s

D
3 s

Solution

We know that acceleration is time derivative of velocity so the given relation can be written as $$\dfrac{dv}{dt}=-5\sqrt[2]{v}$$ or $$\dfrac{dv}{\sqrt[2]{v}}=-5dt$$integrating above equation we get $$2\sqrt[2]{v}=-5t+c$$.....(1)
where c is integrating constant.
To evaluate the constant just put $$t=0$$ and $$v=25$$ as given in the question.
By doing so we lead to $$c=10$$ so the equation-1 becomes $$2\sqrt[2]{v}=-5t+10$$ now put $$v=0$$ in above equation we get $$t=2second$$.
Option C is correct.
Question 8
1 / -0

Two bikes $$A$$ and $$B$$ start from a point. A move uniform speed $$40\ m/s$$ $$B$$ starts from rest with uniform acceleration $$2\ m/s^{2}$$. If $$B$$ starts at $$t = 0$$ and $$A$$ starts from the same point at $$t = 10\ s$$, then the time interval during the journey in which $$A$$ was ahead of $$B$$ is

A
$$20\ s$$

B
$$8\ s$$

C
$$10\ s$$

D
$$A$$ is never ahead of $$B$$

Solution

Let distance covered by A be S1

Let distance covered by B be S2

B will catch A when S1 = S2

S = ut + 1/2 * a * t^2

Therefore,

40t + (1/2 * 0 * t^2) = 0t + 1/2 * 4 * t^2

40t = 2t^2

20t = t^2

Therefore, t = 20
Question 9
1 / -0

A particle $$P$$ starts from the origin with velocity $$\overset{\rightarrow}{u}=(2\hat{i}-4\hat{j})m/s$$ with constant acceleration $$(3\hat{j}-5\hat{j}) m/s^2$$ after travelling for $$2$$ second its distance from the origin is

A
$$10m$$

B
$$20.6m$$

C
$$9.8m$$

D
$$11.7m$$

Solution

$$\vec{s} = \vec{u}t + \dfrac{1}{2}\vec{a}t^2 $$
$$\vec{s} = (2 \hat{i} - 4 \hat{j})2 + \dfrac{1}{2}(3\hat{i} - 5\hat j)(2)^2 $$
$$\vec{s} = (4 \hat{i} - 8 \hat{j}) + (6\hat{i} - 10\hat j) $$
$$\vec{s} = (10 \hat{i} - 18 \hat{j}) $$
$$|\vec{s}| = \sqrt{(10)^2 + (18)^2} $$
$$|\vec{s}| = \sqrt{100 + 324} $$
$$|\vec{s}| = \sqrt{424} $$m =20.6 m
Question 10
1 / -0

A particle starts from rest, accelerates at $$2\ m/s^{2}$$ for $$10s$$ and then goes for constant speed for $$30s$$ and then decelerates at $$4\ m/s^{2}$$ till it stops. What is the distance travelled by it?

A
$$750\ m$$

B
$$800\ m$$

C
$$700\ m$$

D
$$850\ m$$

Solution

A)
Initially particle starts from rest. So$$ u=0 \ and \ a=2m/s^2 and t=10 sec$$ and formula for this case is:
$$s=ut+(1/2)a(t^2)$$ and using this
$$s = 0*10 + (1/2)*2*(10^2) = 100m$$

and using the given information and this formula
v = u + at
we get the value of final velocity and on substituting all values we get:
0+2*10 = 20m/sec
Given that we have constant speed implies initial velocity is the same as the final velocity.

v=u and using this information and the formula $$v=u+at $$
we get a=0 as t can’ t be zero. and $$t=30 sec$$ (given). And from part $$u=20m/sec$$

Now to get the value of distance in this part we use this formula :
$$ s=ut+(1/2)a(t^2)$$ and using this we get:
$$ 20*30+(1/2)*0*(30^2) = 600m .$$

Now $$ a=(-)4m/sec^2$$
as particle is decelerating and given statement is it decelerates till is stops implies that final velocity is zero. And from above part $$u=20m/sec$$ and to get the value of distance we use this formula:
$$ v^2-u^2 = 2as$$ and using this we get:
$$0-20^2 = 2*(-4)*s => s =400/8 = 50m.$$
So using all the part info we get total distance travelled: 100+600+50 = 750

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Motion in A Straight Line Test - 49

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Motion in A Straight Line Test - 49

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Question 1

$$2N$$

$$4 N$$

$$5 N$$

$$1 N$$

Solution

Question 2

$$s_2=s_1$$

$$s_2=2s_1$$

$$s_2=3s_1$$

$$s_2=4s_1$$

Solution

Question 3

$$\dfrac{v^2}{a}$$

$$\dfrac{v^2}{2a}$$

$$\dfrac{2v^2}{a}$$

$$\dfrac{v^2}{3a}$$

Solution

Question 4

(7, 3)

(7, 18)

(21, 7)

(3, 7)

Solution

Question 5

20 m

400 m

1440 m

2880 m

Solution

Question 6

Zero

$$1\ m/s^{2}$$

$$3\ m/s^{2}$$

$$6\ m/s^{2}$$

$$9\ m/s^{2}$$

Solution

Question 7

1 s

5 s

2 s

3 s

Solution

Question 8

$$20\ s$$

$$8\ s$$

$$10\ s$$

$$A$$ is never ahead of $$B$$

Solution

Question 9

$$10m$$

$$20.6m$$

$$9.8m$$

$$11.7m$$

Solution

Question 10

$$750\ m$$

$$800\ m$$

$$700\ m$$

$$850\ m$$

Solution

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