Motion in A Straight Line Test

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Motion in A Straight Line Test - 35

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Question 1
1 / -0

A stone projected up vertically with a velocity v, reaches points x, y and z in its path with velocities $$\displaystyle \frac { v }{ \sqrt { 5 } } ,\frac { v }{ \sqrt { 10 } } and\frac { v }{ \sqrt { 15 } } $$ respectively. Find the ratio xy : yz.

A
3:1

B
4:1

C
6:1

D
3:2

Solution

Let $$S_x$$ , $$S_y$$ and $$S_z$$ are the height from the ground of the point $$x$$, $$y$$ and $$z$$ respectively.
Using $$v^2= u^2+2as$$ $$\Rightarrow (\dfrac{v}{\sqrt5})^2=v^2-2gS_x$$ $$\Rightarrow S_x=\dfrac{v}{25}$$,
$$\Rightarrow (\dfrac{v}{\sqrt10})^2=v^2-2gS_y$$ $$\Rightarrow S_y=\dfrac{9v}{200}$$,
$$\Rightarrow (\dfrac{v}{\sqrt15})^2=v^2-2gS_z$$ $$\Rightarrow S_z=\dfrac{14v}{300}$$,
$$xy=S_y-S_x= \dfrac{9v}{200}- \dfrac{v}{25}=\dfrac{v}{200}$$
$$yz=S_z-S_y= \dfrac{14v}{300}- \dfrac{9v}{200}=\dfrac{v}{600}$$
$$\therefore$$ $$xy:yz=\dfrac{v}{200}:\dfrac{v}{600}=3:1$$
Question 2
1 / -0

A bike moving along a straight road covers $$35$$ m in the $$4$$th second and $$40$$ m in the $$5$$th second. What is its initial velocity and acceleration (if the acceleration is assumed to be uniform )?

A
$$\displaystyle { 17.5\ ms }^{ -1 }$$

B
$$\displaystyle { 8\ ms }^{ -1 }$$

C
$$\displaystyle { 7.8\ ms }^{ -1 }$$

D
$$\displaystyle { 38.5\ ms }^{ -1 }$$

Solution

Distance covered in $$nth$$ second is given by $$S_n = u+\dfrac{1}{2}(2n-1)$$
Case 1 :   $$n = 4$$     $$S_n = 35$$
So,   $$35 = u+\dfrac{a}{2}(2\times 4-1) $$
Or   $$35 = u+3.5 \ n$$           ......(1)
Case 2 :   $$n = 5$$     $$S_n = 40$$
So,   $$40 = u+\dfrac{a}{2}(2\times 5-1) $$
Or   $$40 = u+4.5 \ n$$           ......(2)
Equation (2) - (1), we get $$40-35 = 4.5 a-3.5 a$$
Or   $$a = 5 \ m/s^2$$
Now,    $$35 = u+3.5\times 5$$
$$\implies \ u = 17.5\ m/s$$
Question 3
1 / -0

Find the initial velocity of projection of a ball thrown vertically up if the distance moved by it in 3rd second is twice the distance covered by it in 5th second.$$\displaystyle (Take g = 10\ { ms }^{ 2 })$$

A
$$\displaystyle 45\ { ms }^{ -1 }$$

B
$$\displaystyle 75\ { ms }^{ -1 }$$

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

D
$$\displaystyle 85\ { ms }^{ -1 }$$

Solution

Let initial velocity $$u$$,
Let $$S_1$$ be the distance covered in $$n$$ seconds and $$S_2$$ be the distance covered in $$(n+1)$$.
Let $$S_n$$ be the distance covered in $$nth$$ seconds.
Using $$s=ut+\frac{1}{2}at^2$$, $$\Rightarrow S_1= un-\frac{1}{2}gn^2$$ ....1
$$\Rightarrow S_2= u(n+1)-\frac{1}{2}g(n+1)^2$$ ....2
Subtracting equations 1 from 2, $$S_2-S_1=u(n+1)-\frac{1}{2}gn^2-(un-\frac{1}{2}g(n+1)^2)$$
$$\Rightarrow S_n=S_2-S_1= u-\frac{1}{2}(2n+1)$$
$$\therefore S_3= u-35$$ ......3
$$\therefore S_5= u-55$$ ......4
Using the condition given in question , $$S_3=2S_5$$
$$\Rightarrow u-35=2(u-55)$$ $$\Rightarrow u=75 ms^{-1}$$
Question 4
1 / -0

An object travels 10 s with uniform acceleration along a straight line path. During this period if the velocity of the object is increased from $$\displaystyle { \ 5\ ms }^{ -1 } to { \ 25\ s }^{ -1 }$$, then find the distance travelled by the body?

A
$$150$$ m

B
$$100$$ m

C
$$125$$ m

D
$$175$$ m

Solution

Initial speed $$u = 5 \ m/s$$
Final speed $$v = 25\ m/s$$
Time $$t = 10 \ s$$
Using $$v = u+at$$
$$\therefore$$ $$25 = 5+a\times 10$$
$$\implies \ a = 2 \ m/s^2$$
Using $$S = ut+\dfrac{1}{2}at^2$$
$$S = 5\times 10+\dfrac{1}{2}\times 2\times 10^2 = 150 \ m$$
Question 5
1 / -0

A stone is dropped by a person from the top of a tower, which is 200 m tall. At the same time another stone is thrown upwards, with a velocity of $$\displaystyle 50\ { ms }^{ -1 }$$ by a person standing at the foot of the tower. Find the time after which two stones meet.

A
$$4 \ s$$

B
$$5 \ s$$

C
$$8\ s$$

D
$$10\ s$$

Solution

Let the two stones meet at a distance of x mfrom the top of the tower, and 't' be the time taken. Let us assume the downward direction as positive.
For the stone that is dropped its initial velocity $$ \displaystyle u = 0\quad { ms }^{ -1 }$$ ; displacement s = x and acceleration = acceleration due to gravity (g).
Using $$\displaystyle s=\quad ut+\frac { 1 }{ 2 } { at }^{ 2 },\\ we\quad get\quad x\quad =\quad (0)\quad t\quad +\frac { 1 }{ 2 } { gt }^{ 2 }\quad \rightarrow \quad (1)$$.
For the stone that is projected vertically upwards, its initial velocity, $$\displaystyle u =-50\quad { ms }^{ -1 }$$; displacement s =-(200-x) and acceleration a =g.
using $$\displaystyle s=\quad ut+\frac { 1 }{ 2 } { at }^{ 2 },\\ we\quad get\quad -(200-x)=-50\times t+1/2{ gt }^{ 2 }\\ 200=50t-1/2{ gt }^{ 2 }+x\quad \rightarrow \quad (1)(2)\\ From\quad the\quad equations\quad (1)\quad and\quad (2)\\ we\quad have\quad 200\quad =50t-1/2{ gt }^{ 2 }+1/2{ gt }^{ 2 }\\ \Rightarrow 200=50\therefore \quad t=4s$$
Question 6
1 / -0

Two cars arrive at certain point with velocity of $$\displaystyle { 30\ ms }^{ -1 } , { 25\ ms }^{ -1 } $$ and travel in a straight line with uniform acceleration $$\displaystyle { 0.25\ ms }^{ -2 }\ and\ { 0.5\ ms }^{ -2 } $$ respectively.
(A) Find the distance at which they meet again.
(B) Also determine the time after which the final velocity one of the cars is equal to the initial velocity of the other.

A
$$1400 \ m$$ , $$10\ s$$

B
$$2800 \ m$$ , $$20\ s$$

C
$$1400 \ m$$ , $$30\ s$$

D
$$1000 \ m$$ , $$10\ s$$

Solution

Solution: (A) Both cars travels for same time (t)
First car : $$\displaystyle u= { 30\quad ms }^{ -1 }; a= { 0.25\quad ms }^{ -2 } $$
$$\displaystyle { S }_{ 1 }=(30\times t)+\frac { 1 }{ 2 } (0.25\times { t }^{ 2 })$$
Second car : $$\displaystyle u= { 25\quad ms }^{ -1 }; a= { 0.5\quad ms }^{ -2 } $$
$$\displaystyle { S }_{ 2 }=(25\times t)+\frac { 0.5 }{ 2 }{ t }^{ 2 })$$
$$displaystyle But\quad { S }_{ 1 }={ S }_{ 2 }\\ \therefore 30t+\frac { 1 }{ 2 } { 0.25 }\quad t^{ 2 }\\ =25t+\frac { 1 }{ 2 } { 0.5t }^{ 2 }=10t={ 0.25t }^{ 2 }\\ t=\frac { 10 }{ 0.25 } =40s\\ Length\quad of\quad the\quad path\quad travelled\\ (S)=\quad (30\times 40)+\frac { 0.25 }{ 2 } \times { (40 })^{ 2 }\\ =1200+200=1400m$$
(B) $$\displaystyle v=30\quad { ms }^{ -1 },\quad u=25\quad { ms }^{ -1 }\\ a=0.5\quad { ms }^{ -2 },\quad t=\quad ?\\ v=u+at\Rightarrow 30=25+0.5\times t\\ \Rightarrow 5=0.5t\Rightarrow t=10s$$
Question 7
1 / -0

When a body is dropped from a tower, it covers 75% of total height of the tower in the last second of its fall. What is the total height of the tower?

A
$$4.9$$ m

B
$$ 9.8$$ m

C
$$5.6$$ m

D
$$6.0$$ m

Solution

Let h be the height of the tower the initial velocity u=0
$$\displaystyle h=\frac { 1 }{ 2 } { gt }^{ 2 }\\ { S }_{ n }=\frac { g }{ 2 } (2t-1)=0.75h\\ =\frac { g }{ 2 } (2t-1)=0.75(\frac { 1 }{ 2 } { gt }^{ 2 })\quad (from\quad 1))\\ 2t-1=\frac { 75{ t }^{ 2 } }{ 100 } =\frac { 3 }{ 4 } { t }^{ 2 }\\ 8t-4={ 3t }^{ 2 }\\ { 3t }^{ 2 }-8t+4=0\\ t=\frac { 8\pm \sqrt { 64-4\times 3\times 4 } }{ 2\times 3 } =\frac { 8\pm \sqrt { 16 } }{ 12 } =\frac { 8\pm 4 }{ 12 } \\ t=1,-2/3\Rightarrow t=1s\\ h=\frac { 1 }{ 2 } \times { gt }^{ 2 }=\frac { 1 }{ 2 } \times 9.8\times 1=4.9\quad m$$
Question 8
1 / -0

A freely falling body crosses points P, Q and R with velocity V, $$2$$V and $$3$$V respectively. Find the ratio of the distance PQ to QR.

A
$$5 : 3$$

B
$$3 : 5$$

C
$$1 : 2$$

D
$$2 : 1$$

Solution

Let the acceleration be $$a$$.
For P to Q :
Initial speed $$u = V$$
Final speed $$v = 2V$$
Using $$v^2 - u^2 = 2aS$$
$$\therefore$$ $$2aS_{PQ} = (2V)^2-V^2$$
We get   $$S_{PQ} = \dfrac{3V^2}{2a}$$
For Q to R :
Initial speed $$u = 2V$$
Final speed $$v = 3V$$
Using $$v^2 - u^2 = 2aS$$
$$\therefore$$ $$2aS_{QR} = (3V)^2-(2V)^2$$
We get   $$S_{QR} = \dfrac{5V^2}{2a}$$
So.   $$S_{QR}:S_{PQ} = 3:5$$
Question 9
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A car moves with a constant velocity of $$\displaystyle { 10\ ms }^{ -1 }$$ for 10 s along a straight road. Then it moves with uniform acceleration of $$\displaystyle { 2\ ms }^{ -2 }$$ for $$5$$ seconds. Find the total displacement at the end of the 15 s of its motion.

A
$$175$$ m

B
$$125$$ m

C
$$150$$ m

D
$$105$$ m

Solution

For constant velocity :
Constant velocity $$u = 10 \ m/s$$
Time $$t = 10 \ s$$
So, displacement in 10 seconds $$S_1 = ut = 10\times 10 = 100 \ m$$
For accelerated journey :
Initial speed $$u = 10 \ m/s$$
Acceleration $$a = 2 \ m/s^2$$
Time $$t = 5\ s$$
Displacement in 5 seconds, $$S_2 = ut+\dfrac{1}{2}at^2$$
$$\implies \ S_2 = 10\times 5+\dfrac{1}{2}\times 2\times 5^2 = 75 \ m$$
Total displacement in 15 seconds $$S_{total} = 100+75 = 175 \ m$$
Question 10
1 / -0

A body is dropped from a height of $$2$$ m. It penetrates into the sand on the ground through a distance of $$10$$ cm before coming to rest. What is the retardation of the body in sand?

A
$$\displaystyle -9.8\ {ms} ^{-1}$$

B
$$\displaystyle 196\ {ms} ^{-2}$$

C
$$\displaystyle -196\ {ms} ^{-2}$$

D
$$\displaystyle 9.8\ {ms} ^{-2}$$

Solution

Initial velocity of the body   $$u =0$$
Velocity of the body just before reaching the sand   $$V = \sqrt{2gH}$$
$$\implies \ V = \sqrt{2\times 9.8\times 2} = \sqrt{4\times 9.8} \ m/s$$
Final velocity in the sand $$V_f = 0$$
Distance covered in sand   $$S = 10 \ cm = 0.1 \ m$$
Using   $$V_f^2 - V^2 = 2aS$$
Or   $$0 - (\sqrt{4\times 9.8})^2 = 2\times a\times 0.1$$
$$\implies \ a = -196 \ m/s^2$$
Retardation is $$196 \ m/s^2$$

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 35

Result

Motion in A Straight Line Test - 35

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SHARING IS CARING

Question 1

3:1

4:1

6:1

3:2

Solution

Question 2

$$\displaystyle { 17.5\ ms }^{ -1 }$$

$$\displaystyle { 8\ ms }^{ -1 }$$

$$\displaystyle { 7.8\ ms }^{ -1 }$$

$$\displaystyle { 38.5\ ms }^{ -1 }$$

Solution

Question 3

$$\displaystyle 45\ { ms }^{ -1 }$$

$$\displaystyle 75\ { ms }^{ -1 }$$

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

$$\displaystyle 85\ { ms }^{ -1 }$$

Solution

Question 4

$$150$$ m

$$100$$ m

$$125$$ m

$$175$$ m

Solution

Question 5

$$4 \ s$$

$$5 \ s$$

$$8\ s$$

$$10\ s$$

Solution

Question 6

$$1400 \ m$$ , $$10\ s$$

$$2800 \ m$$ , $$20\ s$$

$$1400 \ m$$ , $$30\ s$$

$$1000 \ m$$ , $$10\ s$$

Solution

Question 7

$$4.9$$ m

$$ 9.8$$ m

$$5.6$$ m

$$6.0$$ m

Solution

Question 8

$$5 : 3$$

$$3 : 5$$

$$1 : 2$$

$$2 : 1$$

Solution

Question 9

$$175$$ m

$$125$$ m

$$150$$ m

$$105$$ m

Solution

Question 10

$$\displaystyle -9.8\ {ms} ^{-1}$$

$$\displaystyle 196\ {ms} ^{-2}$$

$$\displaystyle -196\ {ms} ^{-2}$$

$$\displaystyle 9.8\ {ms} ^{-2}$$

Solution

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