Olympiad Science - Motion
Exam Duration: 45 Mins Total Questions : 30
A particle is travelling with a constant speed. What does this mean
- (a)
The position of the particle remains constant as time passes
- (b)
The particle covers equal distances in equal time intervals
- (c)
The acceleration of the particle is zero
- (d)
The particle does not change its direction of motion.
A particle moving with constant speed covers equal distances in equal time intervals.
A particle shows a value of 46.0 m S-l What can it be?
- (a)
Force of the particle
- (b)
Velocity of the particle
- (c)
Acceleration of the particle
- (d)
Momentum of the particle
A particle with a value of 46.0ms-1represents the velocity of the particle
The positions of a particle moving along a straight line are Xl = 50 m at 10.30 a.m. and x2 = 55 m at 10.35 a.m. respectively. What is the displacement of the particle
between 10.30 a.m. and 10.35 a.m.?
- (a)
2 m
- (b)
5 m
- (c)
7 m
- (d)
9 m
5 = x2 - x1 = 55 - 50 = 5 m
A particle moving with an initial velocity of 5 m S-l is subjected to a uniform acceleration of - 2.5 m S-2. Find the displacement in the next 4 seconds.
- (a)
40 m
- (b)
0 m
- (c)
20 m
- (d)
60 m
\(5=ut+\frac { 1 }{ 2 } at^{ 2 }=(5\times 4)+\frac { 1 }{ 2 } \times (-2.5)\times 16=0\)
An insect moves along a circular path of radius 10 cm with a constant speed. If it takes 1 minute to move from a point on the path to the diametrically opposite point, then what is the distance it travels
and average velocity of the insect respectively?
- (a)
3.14 cm(min)-1,20cm(min)-1
- (b)
3.14 cm(min)-1,0.33cm s-1
- (c)
0.314 cm(min)-1,1 cm(min)-1
- (d)
314 cm(min)-1,0.1 cm(min)-1
Distance\(=\pi r=3.14\times 10=31.4cm\)
Displacement= 2r= 2 x10=20
Average Velocity \(=\frac { s }{ t } =\frac { 20 \ cm }{ 60s } \)
=0.33 cm s-1
If the distance travelled by a body in the nth second is given by (4 + 6 n) m, then find the initial velocity and acceleration of the body respectively.
- (a)
3 m s-1 6 m S-2
- (b)
7m s -1 4 m S-2
- (c)
7 m s -1 6 m S-2
- (d)
7 m s -1 3 m S-2
Sn=(4+6n)..(1)
\({ s }_{ n }=u+\frac { a }{ 2 } (2n-1)\left( u-\frac { a }{ 2 } \right) +an\)....(2)
Comparing (1) and (2) , We get
u= 7 m s-1 and a =6 m s-2
A car moving on a road with uniform acceleration covers 20 m in the first second and 30 m in the next second.
What is its acceleration?
- (a)
20 m S-2
- (b)
10 m S-2
- (c)
30 m S-2
- (d)
5 m S -2
a=\(\frac { { S }_{ 2 }-{ S }_{ 1 } }{ { \triangle t }^{ 2 } } \)=\(\frac { 30-20 }{ { 1 }^{ 2 } } \)=\(\frac { 10 }{ { 1 }^{} } \)m s-2
The velocity of a body is given by the equation v = 6 - 0.02 t, where t is the time taken. What does the body undergo?
- (a)
Uniform retardation of 0.02 m S-2
- (b)
Uniform acceleration of 0.02 m S-2
- (c)
Uniform retardation of - 0.04 m S-2
- (d)
Uniform acceleration of - 0.04 m S-2
Compare the given equation, with v = u + at, we get
u = 6 m s-1 ;a = - 0.02 m S-2
The - sign indicates retardation.
A race car accelerates uniformly from 18.5 m s' to 46.1 m s' in 2.47 seconds. What is the acceleration of the car?
- (a)
11.2 m S-2
- (b)
12.3 m S-2
- (c)
14.5 m S-2
- (d)
16.7 m S-2
a =\(\frac { v-u }{ t } =\frac { 46.1-18.5 }{ 2.47 } =\frac { 27.6 }{ 2.47 } \)=11.2 m S-2
A car covers the first half of the distance between two places at a speed of 40 km h-1 and the second half at 60 km h-1. Find the average speed of the car.
- (a)
100 km h-1
- (b)
50 km h-1
- (c)
48 km h-1
- (d)
52 km h-1
Vavg=\(\frac { 2\times u\times v }{ u+v } =\frac { 2\times 40\times 60 }{ (40+60) } \)=48 km h-1
A particle is moving along a circular track of radius 1 m with uniform speed. Find the ratio of the distance covered and the displacement in half revolution.
- (a)
1: 1
- (b)
0: 1
- (c)
\(\pi\):1
- (d)
\(\pi\):2
In a circular track,
(i) D·tstance covere d in \(\frac { 1 }{ 2 } \)revoIutione=\(\frac { 2\pi{r} }{ 2 } =\pi{r}\)
(iii) Displacement in half revolution = 2r
We have r = 1. \(\therefore \)Ratio = \(\pi\) :2
A body travels 200 cm in the first two seconds and 220 cm in the next four seconds with constant acceleration. Find the velocity at the end of the seventh second from the start.
- (a)
5 cm s-1
- (b)
10 cm s-1
- (c)
15 cm S-1
- (d)
20 cm s-1
We have.S1, = 200 cm; t, = 2 S
S2= 420 cm; t2 = 6 s
s1 = ut1+\(\frac { 1 }{ 2 } \)at12\(\Rightarrow \)u + a = 100 .. (1)
S2 = ut2 +\(\frac { 1 }{ 2 } \)at22\(\Rightarrow \)u + 3a = 70 ..(2)
Solving (1) and (2),we get
a = - 15 cm S-2and u = 115 cm s-1
Since, v = u + at
v = 115 - 15 x 7 = 10 cm s-1
How do the directions of velocity and acceleration act when brakes are applied to a moving cycle?
- (a)
Opposite to each other
- (b)
In the same direction
- (c)
Perpendicular to each other
- (d)
Parallel to each other
When brakes are applied to a moving bicycle, its velocity starts decreasing while it moves in the same direction, i.e., it undergoes retardation. This indicates that its acceleration is in a direction opposite to that of its velocity.
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 in 2 s. Find the ratio of the average speed to the average velocity of the insect.
- (a)
17: 13
- (b)
12: 5
- (c)
13:5
- (d)
17: 12
Ratio =\(\frac { Averagespeed }{ Averagevelocity } =\frac { 17 }{ 13 } \)
A train covers equal displacements in equal intervals of time. Which of the ollowing does it NOT exhibit?
- (a)
Uniform acceleration
- (b)
Uniform speed
- (c)
Uniform velocity
- (d)
None of the above
The body exhibits all the three.
When will a body have zero speed?
- (a)
When a body has uniform acceleration
- (b)
When a body has non-uniform acceleration
- (c)
When a body is always under rest
- (d)
When a body is always under motion
A body under rest always has zero speed and also zero acceleration.
Match the entries in Column-I with those in Column-II.
Column-I | Column-II |
a.Speed | 1.cm |
b.Acceleration | 2.s |
c.Displacement | 3.m s-1 |
d.Time | 4. km h-2 |
- (a)
a -2, b - 1, c - 4, d - 3
- (b)
a - 3, b - 4, c - 1, d - 2
- (c)
a - 4, b - 2, c - 3, d - 1
- (d)
a - 3, b - 2, c - 2, d - 1
The correct sequence is :
a - 3, b - 4, c - 1, d - 2
Speed - m s-1 ,Acceleration - km h-2
Displacement - cm, Time - second(s)
What is the acceleration of a particle moving with uniform velocity?
- (a)
1 m S-2
- (b)
2 m S-2
- (c)
0
- (d)
\(\alpha \)
For a body moving with uniform velocity, change in velocity is zero. 50, acceleration is also zero.
A space shuttle is launched into space. During the first 8 minutes of its launch the average acceleration of the shuttle is 17.5 m S-2.
What is its speed after 8 minutes?
- (a)
8000 m S-1
- (b)
8400 m s-1
- (c)
1200 m s-1
- (d)
1500 m S-1
v = u + at = 0 + 17.5 x (8 x 60)
=8400 m S-1.
A space shuttle is launched into space. During the first 8 minutes of its launch the average acceleration of the shuttle is 17.5 m S-2.
Ram's car does not start, so his friend helps him by pushing it for 10 seconds after which the car reaches a speed of 2 m s-1.
Calculate the acceleration of the car.
- (a)
20 m S-2
- (b)
0.2 m S-2
- (c)
5 m S-2
- (d)
10 m S-2
a =\(\frac { v-u }{ t } =\frac { 2-0 }{ 10 } \)=0.2 m s-1
The two ends of a train moving with uniform acceleration pass a certain point with velocities 6 kmph and 8 kmph respectively. What is the velocity with which the middle point of the train passes the same point?
- (a)
14 kmph
- (b)
5 kmph
- (c)
\(10\sqrt { 2 }\) kmph
- (d)
10 kmph
Let the velocities of the two ends of the train be u and v and the velocity of the middle point be v1
\(\therefore\) The acceleration of the mid-point of the train
a=\(\frac { { { v }_{ 1 }^{ 2 }-u } }{ 2s } =\frac { { v }^{ 2 }-{ v }_{ 1 }^{ 2 } }{ 2s } \)
\(\Rightarrow { v }_{ 1 }=\sqrt { \frac { { v }^{ 2 }-{ u }^{ 2 } }{ 2 } } =\sqrt { \frac { { 6 }^{ 2 }-{ 8 }^{ 2 } }{ 2 } } \)
\(=10\sqrt { 2 } \)kmph
How far does the space shuttle travel in the first 8 minutes?
- (a)
8,000 m
- (b)
80,000 m
- (c)
2,016 km
- (d)
2,600 km
s=ut+\(\frac { 1 }{ 2 } \)at2=0+\(\frac { 1 }{ 2 } \)(17.5)(480)2
= 2016000 m or 2,016 km
What does the area of an 'acceleration displacement' graph represent?
- (a)
Distance
- (b)
Velocity
- (c)
\(\frac { { v }^{ 2 }-{ u }^{ 2 } }{ 2 } \)
- (d)
\(\frac { { v }^{ }-{ u }^{ } }{ t} \)
v2 - u2 = 2aS,\(\therefore\)\(a\times s=\frac { { v }^{ 2 }-{ u }^{ 2 } }{ 2 } \)
(a x S is the area of the accelerationdisplacement graph).
A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110m. Determine the acceleration of the car.
- (a)
6.10 m S-2
- (b)
8.10 m S-2
- (c)
10.10 m S-2
- (d)
12.10 m S -2
S = ut + \(\frac{1}{2}\)at2
110=0 x 5.21 +0.5 x a (5.21)2
=\(\frac{110}{13.57}\)=8.10m s-2
The distance-time graphs of two cyclists moving along a straight line, meet at a point. What can be inferred from this?
- (a)
They collide.
- (b)
They move with the same speed.
- (c)
They are at rest.
- (d)
They start from rest.
If s- t graphs of two cyclists meet at a point, at that time they either pass each other or they collide
A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 12 m s-1, and decreases at the rate of 0.5 m S-2. Find the time it will take to come to rest.
- (a)
6 s
- (b)
12 s
- (c)
24 s
- (d)
48 s
u = 12 m s-1; v = 0; a = - 0.5 m S-2 v = U + at
0= 12 - 0.5 x t , 0.5t = 12, t =\(\frac{12}{0.5}\)=24 s
Sohail cycles on a circular track in anticlockwise direction as shown in the figure. He travels with a speed 'V' to cover the path AB, next with speed '2V' from B to c and with a speed of '3V' from e to A.
What is his average speed for the total journey?
- (a)
2V
- (b)
6V
- (c)
3V
- (d)
\(\frac{v}{2}\)
(i) From A to B,Speed = V
Distance = 2\(\pi\)Rx\(\frac { { 60 }^{ 0 } }{ { 360 }^{ 0 } } \)=\(\frac{2\pi R}{6}\)
Time =\(\frac{D}{S}\)=\(\frac{2\pi R}{6\times v}\)
(ii) From Bto C,Speed = 2V
Distance =2\(\pi\)Rx\(\frac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \)=\(\frac{2\pi R}{3}\)
Time =\(\frac{D}{S}\)=\(\frac{2\pi R}{2\times 3v}\)
(iii) From Cto A, Speed = 3 V
Distance = 2\(\pi\)Rx\(\frac { { 180 }^{ 0 } }{ { 360 }^{ 0 } } \)=\(\frac{2\pi R}{2}\)
Time =\(\frac{D}{S}\)=\(\frac{2\pi R}{2\times 3v}\)
\(\therefore\)Average speed for total journey
=\(\frac{Total Distance}{Tota Itaken}\)=\(\frac { 2\pi R }{ \frac { 2\pi R }{ 6v } +\frac { 2\pi R }{ 6v } +\frac { 2\pi R }{ 6v } } \)
\(=\frac { 2\pi R }{ \frac { 6\pi R }{ 6v } } \)=2v
The motion of an object is plotted in four distance-time graphs. Which of the following graphs given below correctly describes the possible motion of the object?
- (a)
- (b)
- (c)
- (d)
Graph in option (D), shows that the object moves with uniform speed at first and then it decreases (slows down).
A man shot a bullet with a speed of 10m S-1which just penetrates a plank of wood. With what speed should he shoot the bullet so that it passes through 10 similar planks?
- (a)
100 m -1
- (b)
104 m S-1
- (c)
\(10\sqrt { 10 } \) m s-1
- (d)
\(5\sqrt { 10 } \) m s -1
Case (i): S= x, v = 0,u = 10m S-1,
v2- u2= 2aS ,02 - 102= 2ax
-100=2ax=-50=ax ....(1)
Case (ii): S = 10x,v = 0
v2-u2= 2aS, 02- u2= 2a (10x)
-u2 = 20 ax ,= - u2= 20 x-50
u2= 1000 or u =\(10\sqrt { 10 } \) m s -1.
The distance versus time graph of a particle moving is shown below.
What does the graph indicate?
- (a)
The particle starts with certain velocity with retardation and finally comes to rest.
- (b)
The velocity of the particle is constant.
- (c)
The acceleration of the particle is non-uniform throughout.
- (d)
The particle starts with a certain velocity and finally becomes uniform after certain time.
The graph of distance versus time is a curve. Hence, the particle has non-uniform motion. (i.e., it has acceleration).