Motion In A Straight Line

A) The size of the carriage is very small as compared with distance between the stations. So, it can be considered as point object.

B) The size of the monkey is very small as compared with circular track and man so, it can be interpreted as point object.

C) Spinning ball size is very small as compare with ground. Thus, it can be taken as point object.

D) Tumbling beaker is not much smaller than the height of the table to consider it as point object. So, it is not considered as point object.
Hence D is the correct answer

From the given Figure 3.19,

(a) Distance OP < OQ. So, A lives closer to the school than B.

(b) From School, starting time for A is Zero whereas B has some finite value t. So, A starts from the school earlier than B.

(c) Slope represents the velocity in uniform motion.

Slope of B > Slope of A

So, B walks faster than A.

(d) Since, the end point of time for B has less value than A. A and B reaches their homes at different times. B reach home before A.

(e) Only one point of intersection occurs in graph and we know that B is faster than A. So, B overtakes A once in whole journey.

Given,
Speed of the woman = 5 km/h

Distance between her home and office = 2.5 km

Ideal time at office = (12+5) – 9.5 = 7.5 hr.

Speed of the Auto = 25 km/h

Time require to reach the home = 2.5/25 =0.1 h = 6 min

The women will reach the home at 5.06 pm

By interpreting all this data, we obtain graph as follows,

Given,

Distance covered in one step = 1 m

Time required for one step = 1 s

Distance for pit from initial position = 13 m

Time taken to move 5 m forward = 5 s

Time taken to move 3 m backward = 3 s

Speed of the drunkard = 1 m/s (∵from above two observations)

Net distance covered = 5-3 = 2 m

Net time required to cover 2 m = 8 s

Total 4 complete forward-backward cycles and 5 forward steps is possible

Time taken for 4 cycles = 32 sec

Distance covered in 4 cycles = 32 m

∴ Total time required to cover 13 m = 32 +5 = 37 s

The graph obtained from above data is,

Given,

Speed of the jet plane, v_jet = 500 km/h

Relative speed of the combustion products, v’_smoke = 1500km/h

Observation:

As plane moves forward, combustion smoke moves backward.

Thus,

∴ Velocity of combustion gases, v_smoke = v_jet – v’_smoke

= 500 – 1500

= -1000 km/h

(∵ Jet moving direction is taken as ‘+ve’ for man standing on ground)

Given,

Initial velocity of the car, u = 126 km/h

= = 35 m/s

Distance covered before stop, s = 200 m

Final velocity of the car, v = 0 m/s

Let, the retardation be ‘a’.

From the 3^rd equation of motion,

v² – u² = 2as

where,

v = Final velocity

u = Initial velocity

a = Acceleration / Deceleration

s = Distance covered

∴ a = = ms^-2 = -3.06 ms^-2

From the 1^st equation of motion,

v = u + at

where,

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

∴ t = = s = 11.44 s

Given,

The initial velocity of both train A and train B, u = 72 km/h
"

Length of each train, l = 400 m

Acceleration of train A = 0 ms^-2

Acceleration of train B = 1 ms^-2

Time to overtake, t = 50 s

Distance to be covered to overtake,

Let, Distance traveled by train B in "t" time= S_B

Distance traveled by train A in "t" time = SA

We have,

s_B + s_A= initial distance between the trains (d)

From 1^st equation of motion,

v = u + at

where,

v = Final velocity

u = Initial velocity

a = Acceleration/Deceleration

t = Time

Final velocity of train A, v_A = 20 + 0 = 20 m/s

Final velocity of train B, v_B = 20 + (1×50) = 70 m/s

From 2^nd equation of motion,

s = u × t + 0.5 × a × t²

where,

u = Initial velocity

a = Acceleration or Deceleration

s = Distance covered

t = Time

∴
s_A = (20×50) + (0.5×0×50²) = 1000 m

s_B = (20×50) + (0.5×1×50²) = 2250 m

∴ Initial distance between trains, d = 2250 -1000

= 1250 m

Given,

Velocity of car A, v_A = 36 km/h = 10 m/s

Velocity of car B, v_B = 54 km/h = 15 m/s

Velocity of car C, v_C = 54 km/h = 15 m/s

Relative velocity of car B with respect to car A,

v_BA = v_B –(-v_A )= 15-(-10) = 25 m/s

Relative velocity of car C with respect to car A,

v_CA = v_C –v_A= 15-10 =5 m/s

At a certain time, both cars B and C are at the same distance,

s = 1 km = 1000m

Time taken (t) by car B to just reach car A is,

t = = 40 s

∴ The minimum acceleration (a) required for car C to just beat car B in reaching car A is given by,

From 2^nd equation of motion,

s = ut + 0.5at²

where,

u = Initial velocity = 5 m/s

a = Acceleration/Deceleration

s = Distance covered = 1000 m

t = Time = 40 s

⇒ 1000 = (5×40) + (0.5×a×40²)

∴ a = ms^-2 = 1 ms^-2

Hence, car C require minimum acceleration, a = 1 ms^-2 to beat car B.

Given,

Speed of the cyclist, v = 20 km/h = 5.55 m/s

Time period for bus moving in the same direction= 18 min = 1080 s

Time period for bus moving in opposite direction=6 min = 360 s

Let, The velocity if the buses be V.

Thus,

Relative velocity of bus moving in the direction of cyclist = (V-5.55) m/s

Relative velocity of bus moving in opposite direction to the cyclist,

= (V+5.55) m/s

Distance covered by same direction bus = (V-5.55)×1080 m…….(1)

Distance covered by opposite direction bus = (V+5.55)×360 m…..(2)

Since both buses cover same distance (VT), equations (1) and (2) are equal.

⇒ (V-5.55)×1080 m = (V+5.55)×360 m

∴ V = 11.11 m/s

Thus,

Equating, equation (1) = VT

(11.11-5.55)×1080 = 11.11×T

We get, T = 540 s

(a) Irrespective of the direction of the motion of the ball, acceleration due to gravity always acts in the downward direction towards the center of the Earth.

(b) At highest point ball comes to rest. So, velocity of the ball at maximum height is Zero.

(c) During upward motion, Position is negative, Velocity is negative and Acceleration is positive.

During downward motion, Position, velocity and acceleration

all are positive.

(d) Initial velocity of the ball, u = 29.4 m/s

Final velocity of the ball, v = 0 m/s

Acceleration due to gravity, g = 9.8 ms^-2

For freely falling body, 3^rd equation of motion becomes,

v² – u² = -2gs

where,

v = Final velocity

u = Initial velocity

g = Acceleration due to gravity

s = Distance covered

∴ s = m = m = 44.05 m

and, from 1^st equation of motion for freely falling body,

v = u - gt

Where,

v = Final velocity

u = Initial velocity

g = Acceleration due to gravity

t = Time

thus, time of ascent = s = 3s

For freely falling body, time of ascent = time decent

∴ Total time = 2t = 6 s

Hence, the total time taken by the ball to reach players hands = 6 s.

(a) True.

Explanation: When an object is vertically thrown upwards, at maximum height its velocity becomes zero. However, its acceleration remains constant (acceleration due to gravity).

(b) False.

Explanation: Speed and velocity have same magnitude for a body. So, If a body is having zero speed it should have zero velocity.

(c) True.

Explanation: Acceleration is defined as change in velocity with respect to time. Thus, If change in velocity is zero acceleration is also zero.

(d) False.

Explanation: When a body is vertically thrown upwards. Even though positive acceleration the body have it will slowly speeding down due acceleration due to gravity.

True.

Explanation: When a body is freely falling from a height. Acceleration due to gravity speedup the body.

Given,

Ball is dropped from a height = 90 m

Time interval, 0 ≤ t ≤ 12

Initial velocity of the ball =0 m/s

Final velocity of the ball = v m/s

Acceleration due to gravity, g = 9.81 ms^-2

From 2^nd equation of motion for freely falling body,

s = ut + 0.5gt²

where,

u = Initial velocity

g = Acceleration due to gravity

s = Distance covered

t = Time

⇒ 90 = 0 + (0.5×9.81)t²

∴ t = 4.29 s

From 1^st equation of motion for freely falling body,

v = u + gt

where,

v = Final velocity

u = Initial velocity

g = Acceleration due to gravity

t = Time

⇒ v = 0 + (9.81×4.29) = 42.04 m/s

Bounce velocity of the ball, v_b = 0.9v = 37.84 m/s

Time (t’) by the bouncing ball to reach maximum is given by,

v = v_b – gt’

where,

v = Final velocity

v_b = Bounce velocity

g = Acceleration due to gravity

t’ = Bouncing time

⇒ 0 = 37.84 – (9.81×t’)

∴ t’ = 3.86 s

Total time taken by the ball = t + t’ = 4.29 + 3.86 = 8.15 s

The iterations go on like this up to ball reach a static condition.

The graph obtained by the data is,

(a) The shortest path between two points in a travel is defined as Displacement and the actual path traveled is called Distance (or) Total length of the path in same interval.

In above figure, AB is Displacement and AC-CD-DB is distance or Total length of the path in the same time interval T.

(b) Average velocity, v = m/s

Average speed, s = m/s

From the figure,

v = m/s

s = m/s

Given,

Speed from home to market, s₁ = 5 km/h

Speed from market to home, s₂ = 7.5 km/h

Distance between market and home, d = 2.5 km

Average velocity, v = m/s

Average speed, s = m/s

Magnitude of total length of the path = 5 km

Time taken for home to market, t₁ = h = 30 min

Time taken for market to home, t₂ = h= 20 min

Total journey time, T = t₁ + t₂ = 50 min

For,

(i)0 to 30 min

(a) Average velocity = = 5 km/h

(b) Average speed = = 5 km/h

(ii)0 to 50 min

Net displacement = 0 m

Net distance = 5 km

(a) Average velocity = 0 m/s

(b) Average speed = = 6 km/h

(iii)0 to 40 min

Distance travelled in return journey in 10 min = 7.5× =1.25 km

Net displacement = 2.5 – 1.25 km = 1.25 km

Net distance = 2.5+1.25 km = 3.75 km

(a) Average velocity = = 1.875 km/h

(b) Average speed = = 5.625 km/h

Instantaneous velocity is defined as first derivative of distance with respect to time, v_in = m/s

The time interval dt is very small such that direction particle doesn’t change. Since, velocity and speed differ in direction only.

Thus,

Instantaneous velocity = Instantaneous speed ……always

(a) A particle cannot have two positions at the same instant of time in on-dimensional motion. So, it is not one-dimensional motion.

(b) A particle in one dimensional motion never have two velocities at same instant of time. So, it is not one-dimensional motion.

(c) Speed is a scalar quantity so, it cannot be negative. It is not a one-dimensional motion.

(d) Total path length cannot be decrease. So, it is not representing any motion.

No.

x-t graph doesn’t represent the trajectory followed by the particle. It is formed by the points consist of both time and space coordinates.

Given,

Speed of the police van = 30 km/h = 8.33 m/s

Speed of the thief’s car = 192 km/h = 53.33 m/s

Speed of the bullet = 150 m/s

Initial speed of the bullet is given by,

U = Speed of the bullet + Speed of the police van

= 150 + 8.33 = 158.33 m/s

Since, both bullet and thief’s car moving in same direction, relative velocity of the bullet with respect to thief’s car is given by,

V_bt = U – Speed of the thief’s car

= 158.33 – 53.33 = 105 m/s

Hence, The bullet hits the thief’s with a speed of 105 m/s

(a) A ball kicked towards a wall with constant velocity. Initially the ball is at rest and moves with constant velocity when kicked and hits the wall bounds back with certain velocity and comes to rest.

(b) A ball thrown from certain height vertically downwards. Thrown ball hits ground and bounces again and again until it reaches velocity zero (rest).

(c) Hammer moving with constant velocity to hit the nail. Initially acceleration zero (constant velocity) when hammer hits nail it produces jerk for short time and again comes to initial position (acceleration is zero).

The given figure represents simple harmonic motion (SHM).

Acceleration of a particle in simple harmonic motion is given by,

a = -ω²x ………….. (1)

Where,

a = acceleration/deceleration

ω = angular frequency

x = path distance

And v = r ω

Where,

v = velocity

r = radial distance

From above graph,

At t = 0.3 s (lies between t=0 and t=1),

Position, x is Negative,

Velocity, v is Negative (negative slope), and

Acceleration, a is Negative (from equation 1)

At t = 1.2 s (lies between t=1 and t=2),

Position, x is Positive,

Velocity, v is Positive (negative slope), and

Acceleration, a is Negative (from equation 1)

At t = -1.2 s (lies between t=1 and t=2),

Position, x is Negative,

Velocity, v is Positive (negative slope), and

Acceleration, a is Positive (from equation 1)

From the graph,

It is clear that slope of x-t graph is maximum in interval 2 and minimum in interval 3. The interval which covers more lengthy line will have maximum average speed. Thus, average speed is maximum in interval 3.

If slope in x-t graph is positive average speed is positive and slope in x-t graph is negative average speed is also negative.

∴ In interval 1 average speed is positive.

In interval 2 average speed is positive.

In interval 3 average speed is negative.

We know that,

Slope of s-t graph represents acceleration.

y- Coordinate of point on s-t graph represents speed.

From the given s-t graph,

Slope is maximum in interval 2. So, it has maximum average acceleration.

y- Coordinate value of point in interval 3 has highest value thus, interval 3 has maximum average speed.

In interval 1,

Slope is positive. So, acceleration is positive.

y- Coordinate of points are positive. So, speed is positive.

In interval 2,

Slope is negative. So, acceleration is negative.

y- Coordinate of points are positive. So, speed is positive.

In interval 3,

Slope is positive. So, acceleration is positive.

y – Coordinate of points are positive. So, speed is positive.

At A, B and C slopes of curve are ‘0’. Thus, Acceleration at this point are zero.

Straight line. Because,

Distance covered by the body in n^th second is given by the equation,

D_n = u + m

Where,

u = initial velocity,

a = acceleration,

n = time = 1, 2, 3 ……, n

Given,

u = 0 m/s and a = 1 m/s²

∴ D_n = m ……………….. (1)

By substituting n values in equation 1, n = 1, 2, 3……… we get,

The graph obtain by above points is,

It is a straight line not parabola.

Given,

Initial speed of the ball, u = 49 m/s

Acceleration, a = -g = -9.81 m/s²

Speed of the lift = 5 m/s

Case 1, Lift is stationary:

It resembles freely falling body. Thus, In upward motion final velocity, v = 0

From 1^st equation of motion,

v = u + at

Where,

v = Final velocity

u = Initial velocity

a = Acceleration/Deceleration

t = Time

⇒ Time of ascent, t = s = s = 5 s

Total time of float = 2×Time of ascent = 10 s

Case 2, Lift is moving with velocity 5 m/s:

Even though, lift (boy) is moving with velocity 5 m/s the initial velocity also increased by 5 m/s. So, relative velocity of ball with respect to boy remains 49 m/s. Thus, Time of float = 10 s.

Given,

Relative speed of child with respect to belt, v_bc = 9 km/h

Speed of the belt, v = 4 km/h

Distance between parents, d = 50 m

(a) Speed of the child for stationary observer when he moves in same direction of belt is given by, V = v_bc + v

= 9 + 4 = 13 km/h

(b) Speed of the child for stationary observer when he moves in opposite direction of belt is given by, V = v_bc – v

= 9 – 4 = 5 km/h

(c) Since, parents are stationary with respect to belt, child speed is same for both. That is v_bc = 9 km/h = 2.5 m/s

⇒ time taken by child to move towards one of his parents, t =

∴ t = = 20 s

If the motion is viewed in parent’s perspective, answer in (a) and (b) will alter and answer in (C) unaltered. Because, for them child speed is constant and equal to 9 km/h.

For first stone:

Initial velocity, u₁ = 15 m/s

Acceleration, a = -g = -10 m/s²

Using the relation,

x_f = x₀ + ut + 0.5at² ……………..(1)

Where,

x_f = Final position = here x₁, ground, 0 m

x₀ = Initial position = here height of the cliff, 200 m

u = Initial velocity = u₁

a = Acceleration

t = time

⇒ 0 = 200 + 5t + (0.5×10)t²

By solving, t = 8 s

For second stone:

Initial velocity, u₂ = 30 m/s

Acceleration, a = -g = -10 m/s²

Using the relation,

x_f = x₀ + ut + 0.5at² …………..(2)

Where,

x_f = Final position = here x_2, ground, 0 m

x₀ = Initial position = here height of the cliff, 200 m

u = Initial velocity = u₂

a = Acceleration

t = time

⇒ 0 = 200 + 30t + 5t²

By solving, t = 10 s

Subtracting equation 1 from equation 2 to we get,

x₂-x₁ = 15t ………….. (3)

Equation 3 represent straight line.

Due to this linear relation between (x₂-x₁) and t , the remains straight line till 8 s.

So, Maximum separation between two stones occurs at t = 8 s.

∴(x₂-x₁)_max = 15×8 = 120 m

After 8 s, only second stone is in motion whose variation is given by the quadratic equation,

(x₂-x₁) = 200 + 30t + 5t²

Hence, the equations of linear and curved path are given by,

(x₂-x₁) = 15t (linear)

(x₂-x₁) = 200 + 30t + 5t² (curved)

(a) Distance travelled by the particle is equal to the area under s-t graph.

∴ Distance covered in time interval t=0 to t=10 is,

D = 0.5×10×12 m

= 60 m

Average speed = m/s

= = 6 m/s

(b) Since, the particle is undergoing increase in speed in interval t=2 s to t=5 s and decrease in speed in interval t=5 s to t= 6 s. So, we have to deal this interval separately.

Let distance traveled in between 2 s and 5 s be d₁ and distance travelled in between 5 s and 6 s be d₂. Then,

Total distance covered in t=2 s and t=6 s be,

D = d₁ + d₂

For distance d₁:

For time interval t=0 s to t=5 s,

From 1^st equation of motion,

v = u + at

Where,

v = Final velocity = here 12 m/s

u = Initial velocity = here 0 m/s

a = Acceleration/Deceleration = Let a’

t = Time = 5 s

⇒12 = 0 + 5a’

∴ a’ = 2.4 m/s²

Again, from first equation of motion and t = 2 s

v’ = 0 + 2×2.4 = 4.8 m/s

Distance travelled by the particle in t=2 s to t=5s .i.e.,3s

From second equation of motion,

s = ut + 0.5at²

Where,

u = Initial velocity = v’ m/s = 4.8 m/s

a = Acceleration/Deceleration = 2.4 m/s²

s = Distance covered = d₁

t = Time = 3 s

⇒ d₁ = (4.8×3) + (0.5×2.4×3²) = 25.2 m

For distance d₂:

For time interval t=5 s to t=10 s,

Final velocity = velocity at 10 s = 0 m/s

Initial velocity = velocity at 5 s = 12 m/s

From 1^st equation of motion,

v = u + at

Where,

v = Final velocity = here 0 m/s

u = Initial velocity = here 12 m/s

a = Acceleration/Deceleration = Let a’

t = Time = 5 s

⇒ 0 = 12 + 5a’

∴ a’ = -2.4 m/s²

Distance travelled by the particle in t=5 s to t=6 s..i.e.,1s

From second equation of motion,

s = ut + 0.5at²

Where,

u = Initial velocity = v’ m/s = 12 m/s

a = Acceleration/Deceleration = -2.4 m/s²

s = Distance covered = d₁

t = Time = 1 s

⇒ d₂ = (12×1) + (0.5×-2.4×1²) = 10.8 m

∴ Total distance covered in t=2 s and t=6 s be,

D = d₁ + d₂ = 25.2 + 10.8 = 36 m

And, Average speed = m/s

= = 9 m/s

It is clear from the given graph that slope is not linearly varying. All linear equations in the given equations will cancel out and remaining equations may cause the above graph.

Thus,

Equations (a), (b) and (e) are wrong formulae.

Equations (c), (d) and (f) are correct formulae.