Motion Test

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Motion Test - 3

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Question 1
1 / -0

Two trains with lengths 180 m and 220 m respectively, are running in opposite directions with speeds of 40 km/hr and 50 km/hr respectively. The time taken by them in crossing each other will be

A
16 sec

B
17 sec

C
18 sec

D
22 sec

Solution

u = 40 km/hr
v = 50 km/hr
l = 180 m
L = 220 m
Now, we know that speed = distance/time
u+v = l+L / time (When trains move in opposite directions, their speeds are added.)
u+v = 40+50 = 90 km/hr = 90*5/18 = 25 m/s
l+L = 180+220 = 400 m
25 = 400/t
Hence, t = 16 sec
Question 2
1 / -0

The wheel of an engine, with a circumference of 25 decimetres, makes 10 revolutions in 4 seconds. The speed of the wheel (in km/hr) is

A
16

B
22.5

C
32.8

D
36

Solution

Circumference of wheel = 25 dm = 2.5 m
The wheel makes 10 revolutions in 4 s.
Then, in one second = 10/4 = 2.5 revolutions
Speed of wheel = 2.5 x 2.5 = 6.25 m/s
Speed of wheel = 6.25 x 3600/1000 = 22.5 km/hr
Question 3
1 / -0

Two trains running in the same direction at 40 km/hr and 22 km/hr completely pass one another in one minute. If the length of the first train is 125 m, the length of the second train is

A
125 m

B
150 m

C
175 m

D
200 m

Solution

u = 40 km/hr
v = 22 km/hr
l = 125 m
t = 1 min = 60 sec
L = ?
Now, we know that speed = distance/time
u-v = l + L / time (When trains move in same direction, their speeds are subtracted.)
u-v = 40-22 = 18 km/hr = 18*1000/3600 = 5 m/s

5 = 125+L/60
Hence, L = 175 m
Question 4
1 / -0

If a body takes as much time to cover 10 m as a car takes to cover 25 m, the distance the body will cover during the time the car covers 1 km is

A
40 m

B
400 m

C
250 m

D
none of these

Solution

For 25 m of car, body covers = 10 m
For 1 m of car, body covers = 10/25 m
For 1000 m of car, body covers = 10/25 * 1000
= 400 m
Question 5
1 / -0

A train travels the first 15 km at a uniform speed of 30 km/hr, the next 75 km at a uniform speed of 50 km/hr and the last 10 km at a uniform speed of 20 km/hr. Calculate the average speed for the entire journey.

A
100 km/hr

B
40 km/hr

C
75 km/hr

D
20 km/hr

Solution

Time for first 15 km = 15/30 = 0.5 hr
Time for next 75 km = 75/50 = 1.5 hr
Time for next 10 km = 10/20 = 0.5 hr
Total time taken = 2.5 hr
Total distance = 15+75+10 = 100 km

Now, average speed = Total distance travelled / Total time taken
= 100/2.5 = 40 km/hr
Question 6
1 / -0

Gaurav travelled 1200 km by air which formed th of his trip. He travelled one-third of the remaining distance by car and the rest by train. The distance travelled by him in train was

A
480 km

B
800 km

C
1200 km

D
1800 km

Solution

2/5^th of the total distance covered by air = 1200 km
Hence, total distance covered by air = 1200*5/2 = 3000 km
1/3^rd of remaining distance covered by car = 1/3*1800 = 600 km
Distance covered by train = 1800-600 = 1200 km
Question 7
1 / -0

A train overtakes two persons who are walking at the rate of 2 km/hr and 4 km/hr in the same direction in which the train is going, and passes them completely in 9 and 10 seconds respectively. The length of the train (in meters) is

A
54

B
45

C
50

D
72

Solution

To find: Length of train (L)
Speed of train (s) = ?
Train passed first person (at 2 km/hr or 5/9 m/s) in 9 secs.
Train passed second person (at 4 km/hr or 10/9 m/s) in 10 secs.
For 1^st person,
s-5/9 = L/9 (When trains move in the same direction, their speeds are
subtracted.)
or, 9s - L = 5 --------(i)

For 2^nd person,
s-10/9 = L/10
Or 90s - 9L = 100 ---------(ii)

Solving (i) and (ii), we get
Length of train = 50 m
Question 8
1 / -0

The distance between two points A and B is 110 km. A motorcyclist starts from A towards B at 7 a.m. at a speed of 20 km/hr. Another motorcyclist starts from B towards A at 8 a.m. at a speed of 25 km/hr. When will they cross each other?

A
9 am

B
11 am

C
10 am

D
12 am

Solution

Speed of 1^st motorcyclist = 20 km/hr
In 3 hrs or at 10 am, distance covered = 60 km

Speed of 2^nd motorcyclist = 25 km/hr
In 2 hrs or at 10 am, distance covered = 50 km

Total distance covered by both till 10 am = 60+50 = 110 km
Hence, they will cross each other at 10 am.
Question 9
1 / -0

Two trains starting at the same time from two stations, 200 km apart; and going in opposite directions cross each other at a distance of 110 km from one of stations. The ratio of their speeds may be

A
11:20

B
9:20

C
11:9

D
11:10

Solution

Let the distance covered by 1^st train = 110 km
Distance covered by 2^nd train = 90 km
Let the time taken by both = t hrs

Speed of 1^st train : Speed of 2^nd train
110/t : 90/t
11 : 9
Question 10
1 / -0

A person starts his journey from point A and travels 5 m and reaches point B, then takes a 90^o turn and travels another 12 m upto C. What is the total displacement?

A
17 m

B
13 m

C
25 m

D
0 m

Solution

Displacement is the shortest distance.
So, AC² = AB²+ BC²
Hence, AC = 13 m
Question 11
1 / -0

Consider the distance-time graph shown below

During which interval the velocity of the particle is zero?

A
OA

B
AB

C
BC

D
CD

Solution

We know that velocity = displacement/time
In the graph, there is only one interval where displacement is zero.
Hence, velocity of AB = 0
Question 12
1 / -0

Which of the displacement- time graphs is impossible?

A

B

C

D

Solution

We know that time always increases.
In all the graphs, time is increasing, except in 3^rd graph, where the time first increases, then
decreases, which is impossible.
Hence, 3^rd graph is impossible.
Question 13
1 / -0

Consider the following graph

During which interval is the acceleration of the particle positive?

A
OA

B
AB

C
BC

D
CD

Solution

We know that acceleration is positive when distance is positive, i.e. distance increases with time.
In the graph, distance of only one interval increases with time, which is OA.
Hence, OA has the positive acceleration.
Question 14
1 / -0

In the above question, during which interval, is the acceleration of the particle negative?

A
OA

B
AB

C
BC

D
CD

Solution

We know that acceleration is negative when distance is negative, i.e. distance decreases with time.
In the graph, distance of only one interval decreases with time, which is BC.
Hence, BC has negative acceleration.
Question 15
1 / -0

The figure shows the displacement-time graph of a particle moving along x-axis. Which of the following statements will describe the motion of the particle correctly?

A
The particle is moving continuously.

B
The particle moves with a constant velocity upto time t₁ and then stops.

C
The velocity of the particle increases up to time t₁and then becomes constant.

D
The particle is at rest.

Solution

Upto time t₁, the velocity is constant and after t_1, body stops because of no displacement.
Hence, option (2) is correct and option (3) is wrong.
Question 16
1 / -0

The area of a square field is one hectare. The time taken by a boy to run once around it at a speed of 10 km/hr is

A
15.5 minutes

B
10 minutes

C
15 minutes

D
2.4 minutes

Solution

Area of field = 1 hectare = 0.01 sq. km
Side of field = 0.1 km
Total distance covered by boy = 0.1*4 = 0.4 km
Speed of boy = 10 km/hr
Time taken = distance/speed
= 0.4/10 = 0.04 hr = 0.04*60 = 2.4 minutes
Question 17
1 / -0

A column of marching battalion extending to 250 m in length, takes one hour to march through a street at the rate of 50 paces a minute, each pace being of 75 cm. The length of the street is

A
3 km

B
4 km

C
2 km

D
1 km

Solution

Length of battalion (L) = 250 m
Speed (s) = 50 paces/min = 37.5/60 m/s (1 pace = 75 cm = 0.75 m)
(50 paces = 50*0.75 = 37.5 m)
Time (t) = 1hr = 3600 sec
s = (250+L)/t
37.5/60 = (250+L)/3600
Solving this, we get
\'L\' = 2000 m = 2 km
Question 18
1 / -0

A monkey ascends a grease pole 120 meters high, ascending 4 meters in the first minute and slipping down two meters in the next minute, repeating this process till he reaches the top. The time taken by it to reach the top is

A
60 minutes

B
120 minutes

C
117 minutes

D
115 minutes

Solution

In first 2 minutes, height reached after slipping = 4-2 = 2 m
In 4 minutes, height reached after slipping = 4 m
//ly, in 116 minutes, height reached after slipping = 116 m
Now, in the next minute, monkey ascends 4 m and reaches the top.
Hence, total time taken = 117 minutes
Question 19
1 / -0

A body is dropped freely into a river from a bridge at a height of 125 m. Find the time taken by the body to touch the water surface.

A
1 sec

B
3 sec

C
5 sec

D
7 sec

Solution

Time taken to reach surface (t) = ?
Dropping height = 125 m
Initial velocity (u) = 0 m/s (body is at rest initially)
We know that,
s = ut + 1/2gt²
125 = 0+ 1/2*9.8*t²
Solving for t, we get
t = 5.05 sec = 5 sec approx
Question 20
1 / -0

A body is dropped freely into a river from a bridge. It takes 5 s to touch the water surface. The velocity with which, it strikes the water surface is

A
49 m/s

B
15 m/s

C
28 m/s

D
37 m/s

Solution

Time taken to reach surface (t) = 5s
Velocity with which it strikes the surface (v) = ?
Initial velocity (u) = 0 (body is at rest initially)
We know that,
v = u + gt
v = 0 + 9.8*5
v = 49 m/s