Q1: Distinguish between speed and velocity.
Ans: Speed of a body is the distance travelled by a body as per unit time while velocity is the rate and direction of an object’s movement.
Q2: Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?
Ans: If the distance travelled by a body is equal to the displacement, then the magnitude of the average velocity of an object will be equal to its average speed.
Q3: What does the odometer of an automobile measure?
Ans: The odometer of an automobile is used to measure the distance covered by an automobile.
Q4: What does the path of an object look like when it is in uniform motion?
Ans: Graphically the path of an object will be linear; it looks like a straight line when it is in uniform motion.
Q5: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108ms^{−1}
Ans: The given data is that time is five minutes and speed is(3 × 108ms^{−1})
Distance = Speed × Time
⇒ 5min × (3 × 108ms^{−1})
⇒( 5 × 60)sec × (3 × 108ms^{−1})
⇒ 300sec × (3 × 108ms^{−1})
⇒ 900 × 108ms^{−1} = 9 × 10^{10}m
∴ Distance = 9 × 107km
Q6: When will you say a body is in
(i) Uniform acceleration?
Ans: When an object travels in a straight line and its velocity changes by equal amount in an equal interval of time, it is said to have uniform acceleration.
(ii) Nonuniform acceleration?
Ans: Nonuniform acceleration is also called variable acceleration. When the velocity of an object changes by unequal amounts in equal intervals of time, it is said to have nonuniform acceleration.
Q7: A bus decreases its speed from 80kmh^{−1} to 60kmh^{−1} in 5s. Find the acceleration of the bus.
Ans: Initial speed of bus (u) = 80kmh^{−1}
Acceleration (a) =
∴ Acceleration (a) = −1.11m/s^{2}
Q8: What is the nature of the distance time graphs for uniform and nonuniform motion of an object?
Ans: If an object has a uniform motion then the nature of distance time graph will be linear, that is it would in a straight line and if it has nonuniform motion then the nature of the distancetime graph will be a curved line.
Q9: What is the quantity which is measured by the area occupied below the velocitytime graph?
Ans: The area occupied below the velocitytime graph measures the distance moved by any object.
Q10: A bus starting from rest moves with a uniform acceleration of 0.1ms−2 for 2 minutes . Find
(a) The speed acquired,
Ans: u = 0, a = 0.1ms^{−2}, t = 2min = 120sec
v = u + at = 0 + 0.1 × 120=12ms^{−1}
Speed acquired = v = 12ms^{−1}
(b) The distance travelled.
Ans: s = ut + 1/2at^{2} = 0 × 120 + 1/2 x 0.1 × 120^{2 }= 720m
Q11: A trolley, while going down an inclined plane, has an acceleration of2cms^{−2} . What will be its velocity 3s after the start?
Ans: Given : u = 0, a = 2cm/s^{2}, t = 3s
v = u + at = 0 + 2 × 3
= 6cm/s
Q12: A racing car has a uniform acceleration of4ms−2 . What distance will it cover in 10s after start?
Ans: Given: u =0, a = 4m/s^{2}, t = 10s
s = ut + 1/2at^{2}
s = 0 × 10 + 1/2 × 4 × 10^{2}
∴ s = 200m
Q13: Differentiate between distance and displacement?
Ans: The difference between distance and displacement is as below,
Q14: Derive mathematically the first equation of motion V = u + at?
Ans: Acceleration is defined as the rate of change of velocity.
Let V = final velocity; V_{o} = initial velocity, T = time, a = acceleration.
So by definition of acceleration
If V_{o }= u = initial velocity, then [V = u + at]
Q15: alculate the acceleration of a body which starts from rest and travels 87.5m 5sec?
Ans: Given Data: u = 0 (starts from rest) u = initial velocity
a = ? a = acceleration
T = 5sec, t = time
S = 87.5m (S = distance)
From second equation of motion
∴ S = 7m/s^{2} = a
Q16: Define uniform velocity and uniform acceleration?
Ans:
Q17: The velocitytime graph of two bodies A and B traveling along the +x direction are given in the (a) Are the bodies moving with uniform acceleration?
Ans: Yes the bodies are moving with uniform acceleration.
(b) Which body is moving with greater acceleration A or B?
Ans: Body A is moving with greater acceleration.
Q18: Calculate the acceleration and distance of the body moving with 5m/s^{2} which comes to rest after traveling for 6sec?
Ans: Acceleration = a = ?
Final velocity = V = o (body comes to rest)
Distance = s =?
Time = t = 6 sec
From, V = u + at
∴ s = 15m
Q19: A body is moving with a velocity of 12m/s and it comes to rest in 18m, what was the acceleration?
Ans: Initial velocity = u = 12m/s
Find velocity = V = 0
S = distance=18m
A= acceleration =?
From 3^{rd} equation of motion;
Q20: A body starts from rest and moves with a uniform acceleration of 4m/s^{2} until it travels a distance of 800m, find the find velocity?
Ans: Initial velocity =u=0
Final velocity = v = ?
Acceleration=a=4m/s^{2}
Distance = s = 800m
v^{2} − u^{2} = 2as
u^{2} − (0) = 2 × 4 × 800
u = 80m/s
∴ u^{2} = 6400
Q21: Differentiate between scalars and vectors?
Ans: The difference between scalars and vectors is as below,
Q22: An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.
Ans: Yes, if an object is moved through a distance it can have zero displacement because displacement of an object is the actual change in its position when it moves from one position to the other position. So if an object travels from point A to B and then returns back to point A again, the total displacement will be zero.
Q23: A train starting from a railway station and moving with uniform acceleration attains a speed40kmh^{−1} in 10min. Find its acceleration.
Ans: Since the train starts from rest (railway station) = u = zero
Final velocity of train = v = 40kmh^{−1}
time(t) = 10min = 10 × 60 = 600 seconds
Since
Q24:What can you say about the motion of an object whose distance time graph is a straight line parallel to the time axis?
Ans: If the object’s distance time graph is a straight line parallel to the time axis indicates that with increasing time the distance of that object is not increasing hence the object is at rest that is not moving.
Q25: What can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis?
Ans: Such a graph indicates that the object is travelling with uniform velocity.
Q26: A train is travelling at a speed of 90kmh^{−1} . Brakes are applied so as to produce a uniform acceleration of−0.5ms^{−2} . Find how far the train will go before it is brought to rest.
Ans:
Given: a = −0.5ms^{−2}, v = 0 (train is brought to rest)
Q27: A stone is thrown in a vertically upward direction with a velocity of5ms−1 . If the acceleration of the stone during its motion is10ms−1 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Ans: Given: u = 5ms^{−1}, a = −10ms^{−2}
v = 0 (since at maximum height its velocity will be zero)
v = u + at = 5 + (−10) × t
Q28: Two cars A and B are moving along in a straight line. Car A is moving at a speed of 80kmph while car B is moving at a speed of 50kmphin the same direction, find the magnitude and direction of Five v the relative of car A with respect to B.
(a) The relative velocity of car B with respect to A.
Ans: Velocity of car A = 80kmph
Velocity of Car B = − 50 kmph (ve sign indicates that Car B is moving in the opposite direction to Car A )
The relative velocity of car A with respect to B
velocity of car A + ( velocity of car B)
⇒ 80 + (−(−50))
⇒ 80 + 50
⇒ +130kmph
+130kmph
shows that for a person in car B, car A will appear to move in the same direction with a speed of the sum of their individual speed.
(b) Relative velocity of car B with respect to A
Ans:
⇒ Velocity of car B+ ( velocity of car A)
⇒ −50 + (−80)
⇒ −130kmph
It shows that car B will appear to move with 130 kmph in opposite direction to car A
Q29: A ball starts from rest and rolls down 16mdown an inclined plane in 4s.
(a) What is the acceleration of the ball?
Ans: Given: u= initial velocity = 0 (body starts from rest)S= distance = 16 m
T= time = 4s
(b) What is the velocity of the ball at the bottom of the incline?
Ans: From, v = u + at
v = 0 + 2 × 4
[v = 8m/s]
Q30: Two boys A and B, travel along the same path. The displacement – time graph for their journey is given in the following figure.
(a) How far down the road has B travelled when A starts the journey?
Ans: When A starts his journey at4 sec , B has already covered a distance of 857m.
(b) Without calculation, the speed, state who is traveling faster A or B?
Ans: A travels faster than B because A starts his journey late but crosses B and covers more distance then B in the same time as B
(c) What is the speed of A?
Ans:
(d) Are the speed of A and B uniform?
Ans: Yes
(e) What does point X on the graph represent?
Ans: X on the graph represents the point at which both A and B are at the same position
(f) What is the speed of approach of A towards B? What is the speed of separation of A from B?
Ans: Speed of approach of A towards B = 375 m/min−214 m/min
⇒ 161 m/min
Speed of separation of A from B = 161 m/min
Q31: A body is dropped from a height of320m. The acceleration due to the gravity is 10m/s2?
(a) How long does it take to reach the ground?
Ans: Given Data: Height = h
Distance = s = 320m
Acceleration due to gravity = g = 10m/s^{2}
Initial velocity = u = 0
(b) What is the velocity with which it will strike the ground?
Ans: From v = u + at
v = 0 + 10 × 8
v = 80m/s
Q32: Derive third equation of motionv^{2} − u^{2} 2as numerically?
Ans: We know,
When, v= final velocity
u= initial velocity
a = acceleration
t = time
s = distance
From equation (i)
Put the value of t in equation (ii)
Q33: The velocitytime graph of the runner is given in the graph.
(a) What is the total distance covered by the runner in16s?
Ans: We know that area under vt graph gives displacement:
So, Area = distance = s = area of triangle + area of rectangle Area of triangle = 1/2 × base × height
∴Area of triangle = 30m
Area of rectangle= length × breadth
⇒ (16 − 6)×10
⇒ 10 × 10
⇒ 100m
Total area = 180m
Total distance = 180m
(b) What is the acceleration of the runner at t = 11s?
Ans: Since at t = 11sec , particles travel with uniform velocity so, there is no change in velocity hence acceleration = zero.
Q34: A boy throws a stone upward with a velocity of 60m/s .
(a) How long will it take to reach the maximum height(g=−10m/s^{2)} ?
Ans: u = 60 m/s; g = 10m/s^{2}; v=0
The time to reach maximum height is
v = u + at = u + gt
0 = 60 − 10t
t = 60 / 10
=6s
(b) What is the maximum height reached by the ball?
Ans: The maximum height is:
(c) How long will it take to reach the ground?
Ans: The time to reach the top is equal to the time taken to reach back to the ground. Thus, the time to reach the ground after reaching the top is 6s or the time to reach the ground after throwing is 6 + 6 = 12s.
Q35: Derive the third equation of motion − v^{2} − u^{2} = 2as as graphically?
Ans: Let at time t = 0, the body moves with initial velocity u and time at ‘t’ has final velocity ‘v’ and in time ‘t’ covers a distance ‘s’
Put the value of ‘t’ in equation (i)
Third equation of motion.
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