Motion in A Straight Line Test

Result

Motion in A Straight Line Test - 65

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 car drives straight off the edge of a cliff that is $$57 m$$ high. The investigator at the scene of the accident notes that the point of impact is $$130 m$$ from the base of the cliff. How fast was the car traveling when it went over the cliff? (in m/s)

A
$$36$$

B
$$37$$

C
$$38$$

D
$$39$$

Solution

To get the time of flight, used the fact that as the car falls over the cliff the initial vertical component of its velocity is zero.
It then falls under gravity with an acceleration of g;
$$\therefore S=\cfrac { 1 }{ 2 } g{ t }^{ 2 }$$
$$\therefore { t }^{ 2 }=\cfrac { 2s }{ g } $$
$$t=\sqrt { \cfrac { 2s }{ g } } =\sqrt { \cfrac { 2\times 57 }{ 9.8 } } =3.41s$$
Now we know the time of flight, so horizontal component of velocity as it went over the cliff;-
$$v=\cfrac { \triangle s }{ \triangle t } =\cfrac { 130 }{ 3.41 } =38㎧$$
Question 2
1 / -0

A particle $$X$$ moving with a constant velocity $$u$$ crosses a point O. At the same instant another particle $$F$$ starts from rest from O with a constant acceleration $$a$$. The maximum separation between them before they meet is

A
$$u^{2}/2a$$

B
$$u^{2}/a$$

C
$$u/2a$$

D
$$u/a$$

Solution

maximum distance will be when both of them have same velocity because after that point of time velocity of particle F will be more then that of particle X so distance between then starts decreasing.
$$t=\dfrac{u}{a}$$.
distance covered by X till this time is $$S_1=u\times \dfrac{u}{a}$$
distance covered by F till that time is $$s_2=\frac{1}{2}a(\dfrac{u}{a})^2= \dfrac{u^2} {2a}$$
maximum distance between them $$S_1-S_2=\dfrac{u^2}{2a}$$
so option A is correct.
Question 3
1 / -0

A particle is projected upwards with some velocity. At what height from ground should another particle be just dropped at the same time so that both reach the ground simultaneously. Assume that first particle reaches to a maximum height $$H$$.

A
$$3H$$

B
$$4H$$

C
$$5H$$

D
$$6H$$

Solution

Let the height from which the second particle is dropped be $$h$$.
Time of flight of second particle $$T = \sqrt{\dfrac{2h}{g}}$$
Since, both particles hit the ground simultaneously. So both must have equal time of flight.
Time of flight of first particle $$T = \dfrac{2u\sin\theta}{g}$$
Using equation for maximum height $$H = \dfrac{u^2\sin^2\theta}{2g}$$
We get $$H = \dfrac{g}{8}\times T^2 = \dfrac{g}{8}\times \dfrac{2h}{g}$$
$$\implies \ h = 4H$$
Question 4
1 / -0

Directions For Questions

A tortoise and a hare are in a road race to defend the honor of their breed. The tortoise crawls the entire $$1000$$ meters at a speed of $$0.2\ m/s$$. The rabbit runs the first $$200$$ meters at $$2\ m/s$$, stops to take a nap for $$1.3$$ hours, and awakens to finish the last $$800$$ meters with an average speed of $$3\ m/s$$.

...view full instructions

Who wins the race?

A
Tortoise

B
Hare

C
Both reach simultaneously

D
Cannot be judged

Solution

Traveled time of tortoise
$$t=\cfrac { s }{ v } =\cfrac { 1000 }{ 0.2 } =5000s$$
Traveled time of rabbit;-
$$1$$. He runs the first $$200m$$ at $$2㎧$$
$$\therefore { t }_{ 1 }=\cfrac { s }{ v } =\cfrac { 200 }{ 2 } =100s$$
$$2$$. Then he take a nap for $$1.3h$$
$${ t }_{ 2 }=1.3h=1.3\times 3600s$$
$$=4680s$$
$$3$$. Then he run the last $$800m$$ with speed $$3㎧$$
$${ t }_{ 3 }=\cfrac { s }{ v } =\cfrac { 800 }{ 3 } =267s$$
Total traveled time of rabbit
$$={ t }_{ 1 }+{ t }_{ 2 }+{ t }_{ 3 }$$
$$=100+4680+267$$
$$=5047s$$
Since time taken by rabbit is more than time taken by tortoise. Tortoise wins the race by $$47s$$.
Question 5
1 / -0

Directions For Questions

A tortoise can run with a speed of $$10.0 cm/s$$, and a hare can run exactly $$10$$ times as fast. In a race, they both start at the same time, but the hare stops to rest for $$3.00$$ min. The tortoise wins by $$10 cm$$.

...view full instructions

How long does the race take? (in s)

A
$$199.8$$

B
$$180$$

C
$$125.4$$

D
$$190.8$$

Solution

If L is the length of the race then the tortoise run this race with time:
$${ t }_{ tortoise }=\cfrac { L }{ { v }_{ tortoise } } =\cfrac { L }{ 0.1 } \rightarrow 1$$
When the tortoise crosses the finish line the hare is $$20㎝=0.2m$$ behind tortoise. It means that during time $${ t }_{ tortoise }$$ it traveled the distance $$L-0.2$$
Since he stopped for rest of $$3$$ minutes
$$=3\times 603=180s$$
and his speed is
$$10{ v }_{ tortoise }=1㎧$$
$$\therefore L-0.2=10{ v }_{ tortoise }\left( { t }_{ tortoise }-180 \right) $$
Putting $$L=0.1{ t }_{ tortoise }$$ from $$1$$
$$0.1{ t }_{ tortoise }-0.2=1\left( { t }_{ tortoise }-180 \right) $$
$${ t }_{ tortoise }=199.8s$$
Question 6
1 / -0

You are driving along the street at the speed limit ($$35mph$$) and $$50$$ meters before reaching a traffic light you notice it becoming yellow. You accelerate to make the traffic light within the $$3$$ seconds it takes for it to turn red. What is your speed as you cross the intersection? Assume that the acceleration is constant and that there is no air resistance.

A
$$30 mph$$

B
$$40 mph$$

C
$$50 mph$$

D
$$60 mph$$

Solution

This is the motion with constant acceleration. If the acceleration of the car is $$a$$ then we can write the expression for traveled distance.
$$s=ut+\cfrac { 1 }{ 2 } a{ t }^{ 2 }$$
Here,
$$u=$$initial velocity $$=35mph$$
$$=15.645㎧$$
After $$3$$ sec, the car travels distance $$50$$ meters. Then we can find acceleration
$$a=\cfrac { 2(s-{ v }_{ 0 }t) }{ { t }^{ 2 } } $$
$$=\cfrac { 2\left( 50-15.645\times 3 \right) }{ { 3 }^{ 2 } } $$
$$=\cfrac { 2(50-46.935) }{ 9 } $$
$$=0.6811㎨$$
From relation, $$v=u+at$$
$$=15.6+0.68\times 3$$
$$=17.64㎧$$
i.e., $$=40mph$$
Question 7
1 / -0

Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of $$444 m/s$$ in $$1.83$$ seconds, then what is the distance that the sled travels? (in m)

A
$$406 m$$

B
$$40 m$$

C
$$46 m$$

D
$$506 m$$

Solution

Given, initial velocity $$u=0 m/s, $$ final velocity $$v=444 m/s$$ and time $$t=1.83 s$$
If $$a $$ be the acceleration.
Using $$v=u+at$$,
we get $$444=0+a(1.83)$$
$$\implies$$ $$a=444/1.83=242.62\sim 243 m/s^2$$
If $$d$$ be the traveled distance by sled.
Using formula $$v^2-u^2=2aS$$
$$444^2-0^2=2(243)d$$
$$\implies$$ $$d=405.63 \sim 406 m$$
Question 8
1 / -0

Directions For Questions

A rock is shot up vertically upward from the edge of the top of the building. The rock reaches its maximum height $$2 s$$ after being shot. Them, after barely missing the edge of the building as it falls downward, the rock strikes the ground $$8 s$$ after it was launched. Find

...view full instructions

upward velocity the rock was shot at (in m/s)

A
$$19.6$$

B
$$20$$

C
$$21$$

D
$$23$$

Solution

Here free fall motion of the rock. The equations, which describe the motion of the rock are the following;-
$$y(t)={ y }_{ 0 }=+{ v }_{ i }t-9.8\cfrac { { t }^{ 2 } }{ 2 } $$
$$v(t)={ v }_{ i }-9.8t$$
$${ v }^{ 2 }\left( t \right) -{ v }_{ i }^{ 2 }=-2\times 9.8\left[ y(t)-{ y }_{ 0 } \right] $$
where, $${ v }_{ i }=$$initial velocity
$${ y }_{ 0 }=$$initial height (height of the building)
At the maximum height the velocity of the rock is zero,
Then from the second equation we can fine the initial velocity (since $$t=2s$$)
$${ v }_{ i }=9.8t$$
$$=9.8\times 2=19.6㎧$$
Question 9
1 / -0

A body is thrown vertically up from the ground with a speed $$v_{initial}$$ and it reaches maximum height h at time $$t=t_o$$. What is the height to which it would have risen at time $$t=t_o/2$$

A
$$0.75h$$

B
$$h$$

C
$$0.5h$$

D
$$0.3h$$

Solution

In this problem we have free fall motion
$$y(t)={ y }_{ initial }+{ v }_{ initial }t-4.9{ t }^{ 2 }$$
$$v(t)={ v }_{ initial }-9.8t$$
$${ v }^{ 2 }\left( t \right) -{ v }_{ initial }^{ 2 }=-19.6\left[ y(t)-{ y }_{ initial } \right] $$
The initial height of the object is $$0$$ (ground level)
$$\therefore y(t)={ v }_{ initial }t-4.9{ t }^{ 2 }\rightarrow 1$$
$$v(t)={ v }_{ initial }-9.8t\rightarrow 2$$
$${ v }^{ 2 }\left( t \right) -{ v }_{ initial }^{ 2 }=-19.6(t)\rightarrow 3$$
Case I: The object reaches the maximum height h at $${ t }_{ 0 }$$. At the maximum height the velocity of the object is zero and
$$y\left( { t }_{ 0 } \right) =h$$
$$h={ v }_{ initial }{ t }_{ 0 }-4.9{ t }_{ 0 }^{ 2 }\rightarrow 4$$
$$0={ v }_{ initial }-9.8{ t }_{ 0 }\rightarrow 5$$
$${ v }_{ initial }^{ 2 }=19.6h$$
Case II: Height of the object at
$$t=\cfrac { { t }_{ 0 } }{ 2 } $$
Substitute this time in equation $$1$$ and $$3$$
$$y\left( { { t }_{ 0 } }/{ 2 } \right) =\cfrac { { v }_{ initial }{ t }_{ 0 } }{ 2 } =4.9\cfrac { { t }_{ 0 }^{ 2 } }{ 4 } \rightarrow 7$$
$$v\left( { { t }_{ 0 } }/{ 2 } \right) ={ v }_{ initial }-\cfrac { 9.8{ t }_{ 0 } }{ 2 } \rightarrow 8$$
$${ v }^{ 2 }\left( { { t }_{ 0 } }/{ 2 } \right) -{ v }_{ initial }^{ 2 }=-19.6y\left( { { t }_{ 0 } }/{ 2 } \right) \rightarrow 9$$
We know, $${ v }_{ initial }={ t }_{ 0 }9.8$$
Substitute in equation $$4$$ and $$7$$
$$h=9.8{ t }_{ 0 }^{ 2 }-4.9{ t }_{ 0 }^{ 2 }=4.9{ t }_{ 0 }^{ 2 }\rightarrow 10$$
$$y\left( { { t }_{ 0 } }/{ 2 } \right) =\cfrac { 9.8{ t }_{ 0 }^{ 2 } }{ 4 } =4.9\cfrac { { t }_{ 0 }^{ 2 } }{ 4 } $$
$$=3.675{ t }_{ 0 }^{ 2 }\rightarrow 11$$
From equation $$10$$
$${ t }_{ 0 }^{ 2 }=\cfrac { h }{ 4.9 } $$
Substitute this in equation $$11$$
$$y\left( { { t }_{ 0 } }/{ 2 } \right) =3.675{ t }_{ 0 }^{ 2 }$$
$$=3.675\times \cfrac { h }{ 4.9 } $$
$$=0.75h$$
Question 10
1 / -0

From the top of a tower of height $$40m$$, a ball is projected upwards with a speed of $$20m/s$$ at an angle of elevation of $$30^0$$. The ratio of the total time taken by the ball to hit the ground to its time of flight (time taken to come back to the same elevation) is :

A
$$2:1$$

B
$$1:2$$

C
$$1:1$$

D
$$3:2$$

Solution

Time of flight,
$$T=\cfrac { 2v\sin { \theta } }{ g } =\cfrac { 2\times 20\times \sin { 30° } }{ 10 } =2sec$$
Total time taken by the ball to hit the ground is
$$t=\cfrac { distance }{ velocity } =\cfrac { 40 }{ 20\sin { 30° } } =\cfrac { 40 }{ 20\times \cfrac { 1 }{ 2 } } =4sec$$
So, the ratio is $$4:2=2:1$$

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 - 65

Result

Motion in A Straight Line Test - 65

Score

-

Rank

-

SHARING IS CARING

Question 1

$$36$$

$$37$$

$$38$$

$$39$$

Solution

Question 2

$$u^{2}/2a$$

$$u^{2}/a$$

$$u/2a$$

$$u/a$$

Solution

Question 3

$$3H$$

$$4H$$

$$5H$$

$$6H$$

Solution

Question 4

Tortoise

Hare

Both reach simultaneously

Cannot be judged

Solution

Question 5

$$199.8$$

$$180$$

$$125.4$$

$$190.8$$

Solution

Question 6

$$30 mph$$

$$40 mph$$

$$50 mph$$

$$60 mph$$

Solution

Question 7

$$406 m$$

$$40 m$$

$$46 m$$

$$506 m$$

Solution

Question 8

$$19.6$$

$$20$$

$$21$$

$$23$$

Solution

Question 9

$$0.75h$$

$$h$$

$$0.5h$$

$$0.3h$$

Solution

Question 10

$$2:1$$

$$1:2$$

$$1:1$$

$$3:2$$

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.