Motion in A Straight Line Test

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Motion in A Straight Line Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

A cart is released from rest at the top of a long ramp at time $$t = 0$$ seconds and moves down the ramp at a constant acceleration rate. At a time of t, the cart has reached a speed of $$2 m/s$$. How fast will the cart be moving at the time of $$3t$$?

A
$$3 m/s$$

B
$$6 m/s$$

C
$$12 m/s$$

D
$$18 m/s$$

Solution

Initial speed of the cart $$u = 0$$ m/s
Speed of cart after time $$t$$ $$v = 2$$ m/s
Using $$v = u+at$$
$$\therefore$$ $$2 = 0+ at$$ $$\implies at = 2$$

Speed of cart at the instant $$3t$$ $$v' = u + at'$$ where $$t' = 3t$$
$$\therefore$$ $$v' =0+ a(3t) = 0+ 3 \times 2 = 6$$ m/s
Question 2
1 / -0

A 10 kg object free-falling from a cliff. Find out the velocity of the object after 1 sec and after 2 sec?

A
9.8 m/sec after 1 sec and 19.6 m/sec after 2 sec

B
9.8 m/sec after 1 and 2 sec

C
9.8 m/sec after 1 sec and 29.4 m/sec after 2 sec

D
19.6 m/sec after 1 sec and 29.4 m/sec after 2 sec

E
Cannot determine with information provided

Solution

Given : $$m =10 kg$$ $$a = g = 9.8 m/s^2$$ $$u = 0$$ m/s
Using $$v = u+at$$
Velocity of object after 1 s $$v_1 =0+ 9.8 \times 1 =9.8 $$ $$m/s$$
Velocity of object after 2 s $$v_2 =0+ 9.8 \times 2 =19.6 $$ $$m/s$$
Question 3
1 / -0

A cat running with constant acceleration in one direction is at rest initially and has velocity $$v\left(t\right)=1.00{m}/{s}$$ at $$2.00$$ seconds later.
At $$t=4.00$$ seconds, how far away from the starting point will the cat be?

A
$$0.00m$$

B
$$1.00m$$

C
$$2.00m$$

D
$$4.00m$$

E
$$8.00m$$

Solution

Initial speed of the cat $$u = 0 $$ m/s
Speed of cat after $$t = 2$$ s $$v = 1.00$$ m/s
Let the acceleration of cat be $$a$$.
Using $$v = u+at$$ where $$t =2$$ s
$$\therefore$$ $$1.00 = 0 + a \times 2 $$ $$\implies a = 0.5$$ $$m/s^2$$

Thus distance covered by cat in 4 s, $$S = ut'+ \dfrac{1}{2}a(t')^2$$ where $$t' = 4$$ s
$$\therefore$$ $$S = 0+ \dfrac{1}{2} \times 0.5 \times 4^2 = 4.00$$ m
Question 4
1 / -0

A driver applies brakes when he sees a child on the railway track, the speed of the train reduces from 54 km $$h^{-1}$$ to 18 km $$h^{-1}$$ in 5 s. What is the distance travelled by the train during this interval of time?

A
52 m

B
50 m

C
25 m

D
80 m

Solution

Initial velocity , $$u=54 km/h=54(1000/3600) m/s=15 m/s$$ and final velocity, $$v=18 km/h=18(1000/3600)=5 m/s$$
Using formula, $$v=u-at$$, $$5=15-a(5) $$ or $$a=2 m/s^2$$, this is deacceleration of train.
Using, $$S=ut-\dfrac{1}{2}at^2, $$ the distance traveled by train, $$d=15(5)-0.5(2)(5)^2=50 m$$
Question 5
1 / -0

A particular rocket is propelled in one direction such that its acceleration as a function of time is expressed by $$a\left(t\right)=A{t}^{2}+B$$ where $$A$$ and $$B$$ are constants.
The position function of the rocket should depend on which power of $$t$$?

A
$${t}^{1}$$

B
$${t}^{2}$$

C
$${t}^{3}$$

D
$${t}^{4}$$

E
Not enough information

Solution

The acceleration of the rocket $$a(t) =At^2 + B$$
Integrating w.r.t time we get $$v(t) = \int a(t) dt = \int (At^2 + B) dt$$
$$\therefore$$ $$v(t) = \dfrac{A}{3}t^3 + Bt + C$$

Integrating again w.r.t time we get $$x(t) =$$ $$\int v(t) dt = \dfrac{A}{12} t^4 + \dfrac{B}{2}t^2 + Ct + D$$
Thus the position function of the rocket depends on the fourth power of time i.e $$t^4$$
Question 6
1 / -0

A $$50.0\ kg$$ boy is sitting on an amusement park ride where he accelerates straight upward from rest to a speed of $$30.0 m/s$$ in $$3.0 s$$
What is his mass as he accelerates upward?

A
$$990.0 kg$$

B
$$100.0 kg$$

C
$$50.0 kg$$

D
$$5.00 kg$$

E
$$0 kg$$

Solution

The mass of a body is a universal constant which does not change with acceleration. It is a property of the object itself.
Hence the mass remains the same and equal to $$50.0kg$$
Question 7
1 / -0

Two trains depart from one station, one going north at $$30.00$$ miles per hour, and another going west, steadily accelerating with a rate of $$0.3333$$ miles per minute.
How many minutes after departure would the two trains be $$50.00$$ miles apart?

A
$$17.19min$$

B
$$76.28min$$

C
$$89.54min$$

D
$$120.0min$$

E
$$240.0min$$

Solution

Speed of the train going north is given by
$${ v }_{ 1 }=\frac { 30 }{ 60 } miles/minute\\ \quad \quad =0.5miles/min$$
Let $$t$$ be the time when both trains are 50 miles apart, then the distance covered by northward going train is,
$${ d }_{ 1 }={ v }_{ 1 }t\\ \quad =0.5t\quad miles$$
Displacement of train going westwards at time $$t$$,
$${ d }_{ 2 }=\frac { 1 }{ 2 } a{ t }^{ 2 }miles\\ \quad \quad =\frac { 1 }{ 2 } \times 0.3333{ t }^{ 2 }miles\\ \quad \quad =0.1666{ t }^{ 2 }miles$$
Now since the angle between the directions of both trains is $${ 90 }^{ \circ }$$, the distance between them at time $$t$$ is given by
$${ d }^{ 2 }={ { d }_{ 1 } }^{ 2 }+{ { d }_{ 2 } }^{ 2 }\\ { 50 }^{ 2 }={ (0.5t) }^{ 2 }+{ (0.1666{ t }^{ 2 }) }^{ 2 }\\ Let\quad { t }^{ 2 }be\quad x,\\ 0.02775{ x }^{ 2 }+0.25x=2500\\ 0.00111{ x }^{ 2 }+0.01x=100\\ Solving\quad the\quad quadratic\quad equation,\quad we\quad get\\ x=295.6\\ Thus,\quad t=\sqrt { 295.6 } =17.19min$$
Question 8
1 / -0

At an airport, a bored child starts to walk backwards on a moving platform. The child accelerates relative to the platform with $$a=-0.5{m}/{{s}^{2}}$$ relative to the platform.
The platform moves with a constant speed $$v=+1.0{m}/{s}$$ relative to the stationary floor.
In $$4.0$$ seconds, how much will the child have been displaced relative to the floor?

A
$$8m$$

B
$$4m$$

C
$$3m$$

D
$$0m$$

E
$$-4m$$

Solution

The distance the platform displaces with respect to floor in the given time=$$vt=+1m/s\times 4s=+4m$$
The distance the child walks on the platform with respect to it=$$\dfrac{1}{2}at^2=-4m$$
Hence net displacement of child with respect to floor=$$4m+(-4m)=0m$$
Question 9
1 / -0

The length of a minute hand of a clock is 4 cm. Find the displacement and average velocity of the tip of the minute hand when it moves during a time interval from
(a) 3: 15 pm to 3 : 30 pm(b) 4: 15 pm to 4: 45 pm.

A
(a) $$\dfrac{\sqrt{2}}{225} cm s^{-1}$$ (b) $$\dfrac{2}{225} cm s^{-1}$$

B
(a) $$\dfrac{\sqrt{3}}{225} cm s^{-1}$$ (b) $$\dfrac{1}{225} cm s^{-1}$$

C
(a) $$\dfrac{\sqrt{2}}{225} cm s^{-1}$$ (b) $$\dfrac{1}{225} cm s^{-1}$$

D
(a) $$\dfrac{\sqrt{12}}{225} cm s^{-1}$$ (b) $$\dfrac{2}{225} cm s^{-1}$$

Solution

Distance travelled$$=(4^2+4^2)^{1/2}=4\sqrt 2$$
Time taken$$=15\times 60=900s$$
$$v=\cfrac{4\sqrt2}{900}=\cfrac{\sqrt 2}{225}m/s$$
Total displacement $$=8cm$$
Time taken$$=30\times 60=1800s$$
$$v=\cfrac{8}{1800}=\cfrac{1}{225}m/s$$
Question 10
1 / -0

An insect moves along the sides of a wall of dimensions 12 m x 5 m starting from one corner and reaches the diagonally opposite corner. If the insect takes 2 s for its motion then find the ratio of average speed to average velocity of insect.

A
17 : 13

B
27 : 13

C
17 : 16

D
cant say

Solution

$$AB$$ displacement, $$t=2s$$
And $$(AC+BC)$$ is the distance
$$AB=\sqrt{5^2+12^2}=13m$$
$$v=\cfrac{13}{2}m/s$$
Average distance$$=12+5=17m$$
Average speed$$=\cfrac{17}{2}m/s$$
Average speed:Average velocity$$=\cfrac{(17/2)}{(13/2)}=\cfrac{17}{13}=17:13$$

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 38

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Motion in A Straight Line Test - 38

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Question 1

$$3 m/s$$

$$6 m/s$$

$$12 m/s$$

$$18 m/s$$

Solution

Question 2

9.8 m/sec after 1 sec and 19.6 m/sec after 2 sec

9.8 m/sec after 1 and 2 sec

9.8 m/sec after 1 sec and 29.4 m/sec after 2 sec

19.6 m/sec after 1 sec and 29.4 m/sec after 2 sec

Cannot determine with information provided

Solution

Question 3

$$0.00m$$

$$1.00m$$

$$2.00m$$

$$4.00m$$

$$8.00m$$

Solution

Question 4

52 m

50 m

25 m

80 m

Solution

Question 5

$${t}^{1}$$

$${t}^{2}$$

$${t}^{3}$$

$${t}^{4}$$

Not enough information

Solution

Question 6

$$990.0 kg$$

$$100.0 kg$$

$$50.0 kg$$

$$5.00 kg$$

$$0 kg$$

Solution

Question 7

$$17.19min$$

$$76.28min$$

$$89.54min$$

$$120.0min$$

$$240.0min$$

Solution

Question 8

$$8m$$

$$4m$$

$$3m$$

$$0m$$

$$-4m$$

Solution

Question 9

(a) $$\dfrac{\sqrt{2}}{225} cm s^{-1}$$ (b) $$\dfrac{2}{225} cm s^{-1}$$

(a) $$\dfrac{\sqrt{3}}{225} cm s^{-1}$$ (b) $$\dfrac{1}{225} cm s^{-1}$$

(a) $$\dfrac{\sqrt{2}}{225} cm s^{-1}$$ (b) $$\dfrac{1}{225} cm s^{-1}$$

(a) $$\dfrac{\sqrt{12}}{225} cm s^{-1}$$ (b) $$\dfrac{2}{225} cm s^{-1}$$

Solution

Question 10

17 : 13

27 : 13

17 : 16

cant say

Solution

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