Motion | Numericals for Class 11 Students

1. An old person moves on a semi-circular track of radius 40.0 m during a morning walk. If he starts at one end of the track and reaches at other end, find the distance covered and displacement of the person.

In this question, it’s given that person is walking on a semi-circular track of radius 40.0 m
Let’s first draw this situation as a diagram.

Man walking on semi-circular path of radius 40.0 m

So person is going from one end of semi-circular track and then we need to find out distance and displacement covered by him.

For distance as he’s going along circumference so distance covered will be equal to circumference of this semi-circle.

Circumference of a semi-circle = 𝛑r
Where r = radius of semi-circle

As per question, radius of semi-circle is given as 40.0 m
r = 40.0 m

Circumference of semi-circle = 𝛑 × 40.0

Replacing 𝛑 = 3.14

Circumference of semi-circle = 3.14 × 40.0 = 125.6 m

Therefore distance covered by man from one end of semi-circle to another is 125.6 m

Let’s now find out displacement?

Man walking on semi-circular path of radius 40.0 m

From the diagram it’s clear that actual difference between positions of man from starting point of semi-circle on one end to another end is just length 2r.

Displacement of man from one end of semi-circle to another end of semi-circle

Therefore displacement of man will be 2r = 2 × 40.0 = 80.0 m


2. The distance travelled by a particle in time t is given by s = (2.5 m/s2) t2
Find
a) The average speed of the particle during time 0 to 5.0 second
b) The instantaneous speed at t = 5.0 second

Let’s solve each part
Solution for a)

The average speed of a particle in a time interval is defined as the distance travelled by the particle divided by the time interval. If particle travels a distance s in time t1 to t2, the average speed is defined as

\begin{equation} v_{av} = \frac {s} {t_{2} – t_{1}} \end{equation}

In the current question, we need to find out Average Speed of particle during time 0 to 5.0 second

Total Distance covered by particle during this time interval = s = 2.5 × (5.0)2 = 2.5 × 25.0 = 62.5 m

Total time taken to cover this distance = 5 – 0 = 5 second

⇒ Average Speed = Total Distance/Time Taken = 62.5/5 = 12.5 m/s

Therefore if distance travelled by a particle in time t is given as s = (2.5 m/s2) t2 then average speed of the particle during time 0 to 5.0 second is 12.5 m/s

Solution for b)
Instantaneous Speed of a particle is defined as speed of a particle at a particular instance of time.
Mathematically this is defined as

\begin{equation} v=\lim _{\Delta t \rightarrow 0} \frac{\Delta s}{\Delta t}=\frac{d s}{d t} \end{equation}

Where
s = Distance function
t = Time

In the question, its given that s = (2.5 m/s2) t2

So

\begin{equation} v=\lim _{\Delta t \rightarrow 0} \frac{\Delta s}{\Delta t}=\frac{d s}{d t} = \frac{d(2.5 t^{2})}{dt} \\ = 2.5 \frac{d{t^{2}}}{dt} = 2.5 × 2 t = 5.0 t \\ \implies \text{Instantaneous Speed} = 5.0t \end{equation}

Therefore instantaneous speed of this particle is 5.0t

Thus at particular instant t = 5.0 second

Instantaneous Speed = 5.0 × 5.0 = 25.0 m/s

Therefore if distance travelled by a particle in time t is given as s = (2.5 m/s2) t2 then instantaneous speed of the particle at instant t = 5.0 second is 25.0 m/s


3. Below diagram shows speed versus time graph for a particle. Find the distance travelled by the particle during the time t = 0 to t = 3 second.

Distance travelled in time interval t1 to t2 = Area of velocity-time graph from t1 to t2

From the diagram above it’s clear that area of velocity-time graph from t = 0 to t = 3 is a area of right angled triangle which have
base = 3
height = 6

Therefore distance travelled = 0.5 × base × height = 0.5 × 3 × 6 = 9 m

So distance travelled by particle from time t = 0 to t = 3 is 9 m.


4. A particle starts with an initial velocity 2.5 m/s along the positive x direction and it accelerates uniformly at the rate 0.50 m/s2

a) Find the distance travelled by it in the first two seconds

b) How much time does it take to reach velocity 7.5 m/s

c) How much distance will it cover in reaching the velocity 7.5 m/s

Information given in the question

Initial velocityu2.5 m/s
Accelerationa0.50 m/s2

Let’s solve each part one by one

Solution for a)
Find the distance travelled by it in the first two seconds

\begin{equation} x = ut + \frac{1}{2}at^{2} \end{equation}

Replacing
u = 2.5 m/s
a = 0.50 m/s2
t = 2 second

\begin{equation} x = 2.5 × 2 + \frac{1}{2} × 0.50 × 2^{2} \\ x = 5.0 + \frac{1}{2} × 0.50 × 4 \\ x = 5.0 + \frac{1}{2} × 2.00 \\ x = 5.0 + 1.00 = 6 m \end{equation}

Therefore if initial velocity of a particle is 2.5 m/s and its accelerating at rate of 0.50 m/s2 then distance travelled by it in first 2 seconds of it’s motion is 6 metres.

Solution for b)
How much time does it take to reach velocity 7.5 m/s

\begin{equation} v = u + at \end{equation}

Replacing
v = 7.5 m/s
u = 2.5 m/s
a = 0.50 m/s2
t = ????

We need to figure out how much time does particle takes to reach velocity of 7.5 m/s if its initial velocity is 2.5 m/s and it’s accelerating at rate 0.50 m/s2.

\begin{equation} 7.5 = 2.5 + 0.50 × t \\ 7.5 – 2.5 = 0.50t \\ 5.0 = 0.50t \\ \frac{5.0}{0.50} = t \\ t = 10 seconds \end{equation}

Therefore if a particle starts moving with speed 2.5 m/s and is accelerating at rate 0.50 m/s2 then it will take 10 seconds to reach velocity 7.5 m/s.

Solution for c)
How much distance will it cover in reaching the velocity 7.5 m/s

\begin{equation} v^{2} – u^{2} = 2ax \\ \end{equation}

Replacing
v = 7.5 m/s
u = 2.5 m/s
a = 0.50 m/s2
x = ???

\begin{equation} (7.5)^2 – (2.5)^2 = 2 × 0.50 × x \\ 56.25 – 6.25 = 1.00x \\ 50 = 1.00x \\ x = 50 metre \end{equation}

Therefore if a particle starts moving with speed 2.5 m/s and is accelerating at rate 0.50 m/s2 then it will travel 50 metres to reach velocity 7.5 m/s.

Below is a summary of all the answers for this question

A particle starts with an initial velocity 2.5 m/s along the positive x direction and it accelerates uniformly at the rate 0.50 m/s2
Find the distance travelled by it in the first two seconds6 metres
How much time does it take to reach velocity 7.5 m/s10 seconds
How much distance will it cover in reaching the velocity 7.5 m/s50 metres

5. A particle having initial velocity u moves with a constant acceleration a for a time t.

a) Find the displacement of the particle in last 1 second

b) Evaluate it for u = 5 m/s, a = 2 m/s2 and t = 10 second.

Solution for part a)
Find the displacement of the particle in last 1 second
It’s given that
Initial velocity = u
Acceleration = a
Time = t

We need to find out what’s displacement of the particle in last 1 second

This should be equal to displacement of particle in time t minus displacement of particle in time (t – 1)

Let’s suppose that

Dt = Displacement of particle in time t
Dt -1 = Displacement of particle in time (t – 1)

Displacement of particle in last 1 second = Dt – Dt -1

Therefore if initial velocity of a particle is u and it’s moving with constant acceleration a for time t, then it’s displacement in last 1 second of it’s motion will be u + a(2t – 1)/2

Solution for part b)
Evaluate it for u = 5 m/s, a = 2 m/s2 and t = 10 second

This means we need to find out displacement of particle in last 1 second of it’s motion, it’s initial velocity is 5 m/s and it’s constant acceleration is 2 m/s2 for time 10 seconds.

Using the formula which we derived in part a

Displacement of particle in last 1 second = u + a(2t – 1)/2

Replacing
u = 5 m/s
a = 2 m/s2
t = 10 second

Displacement of particle in last 1 second
= 5 + 2(2 × 10 – 1)/2
= 5 + (20 – 1)
= 5 + 19 = 24 m

Therefore if a particle is moving starts moving with initial velocity 5 m/s and is moving at constant acceleration of 2 m/s2 for 10 seconds then it’s displacement in last 1 second of its motion will be 24 metres.


6. A ball is thrown up at a speed of 4.0 m/s
Find the maximum height reached by the ball.

This numerical is quite simple and easy to solve. Just need to use third equation of motion.

v2 – u2 = 2ay

u = 4.0 m/s

v = 0 (At maximum height of it’s motion, velocity of ball will become zero)

a = – 10 m/s2 (because direction of motion of ball is opposite to direction of force of gravity)

y = We need to calculate this

Putting in all these values into equation v2 – u2 = 2ay

02 – (4.0)2 = 2 × (-10) × y

– 16 = – 20y

16/20 = y

0.8 m = y

Therefore if a ball is thrown up at a speed of 4.0 m/s then maximum height reached by it will be 0.8 metres from surface of ground.


7. A particle moves in X-Y plane with a constant acceleration of 1.5 m/s2 in the direction making an angle of 37° with the X-axis.

At t = 0 the particle is at the origin and its velocity is 8.0 m/s along the X-axis.

Find the velocity and the position of the particle at t = 4.0 second.

XY graph showing a particle moving with velocity 8.0 m/s along X-axis and having an acceleration 1.5 metre per second squared at an angle of 37 degrees with X-axis

In this question we need to find out what’s velocity and position of the particle at time t = 4.0 second.

As the particle is moving in XY plane thus both velocity and it’s position must have both x and y components at any particular instant of time.

So if we can figure out what’s X and Y components of velocity, position of particle at time t = 4.0 second then we can easily find out it’s velocity and position by combining together X and Y components.

As per question its given that

ux = 8.0 m/s

a = 1.5 m/s2

Let’s figure out X and Y components of velocity of particle at time t = 4.0 second

Using First Equation of Motion
vx = ux + axt

Replacing
ux = 8.0 m/s

t = 4.0 s

ax = a Cos37° (Just X component of acceleration affects velocity along X axis)

Value of Cos37° = 0.7986
a = 1.5 m/s2

ax = 1.5 × 0.7986 = 1.1979

ax = 1.1979 m/s2

Putting these values in equation vx = ux + axt

vx = 8.0 + 1.1979 × 4.0
= 8.0 + 4.7916
= 12.8

vx = 12.8 m/s

Therefore at t = 4.0 second, X component of velocity of particle is 12.7916 m/s

Let’s now find out what’s Y component of velocity of particle at t = 4.0 second.

Similarly to vx

vy can be calculated using First Equation of Motion vy = uy + ayt

uy = 0 (From the graph above it’s clear that particle have initial velocity only along X axis)

t = 4.0 second

ay = a Sin37° (Value of acceleration only along Y axis matter to calculate particle’s velocity along Y axis)

ay = 1.5 × 0.60181 = 0.902715

ay = 0.902715 m/s2

Placing these values in equation vy = uy + ayt

vy = 0 + 0.902715 × 4.0 = 3.61

vy = 3.61 m/s

Therefore at time t = 4.0 second X and Y components of velocity of particle are 12.8 and 3.61 m/s respectively.

So at t = 4.0 second

vx = 12.8 m/s
vy = 3.61 m/s

\begin{equation} v = \sqrt{v_{x}^{2} + v_{y}^{2}} \\ v = \sqrt{(12.8)^2 + (3.61)^2} \\ v = \sqrt{163.84 + 13.03} \\ v = \sqrt{176.87} = 13.3 \\ v = 13.3 \text{ m/s} \end{equation}

Therefore overall velocity of particle at time t = 4.0 second is 13.3 m/s but as particle is moving in XY plane. So it’s important to also figure out what’s direction of this velocity.

\begin{equation} Tanθ = \frac{v_{y}}{v_{x}} \\ Tanθ = \frac{3.61}{12.8} = 0.282 \\ Tanθ = 0.282 \\ θ = Tan^{-1}\left(0.282\right) \\ θ = 15.74° \end{equation}

Therefore at time t = 4.0 second, velocity of particle is 13.3 m/s and it’s at an angle 15.74° with respect to X axis.

Let’s now find out what’s position of particle at time t = 4.0 second in XY plane.

Let’s first find out X component of particle’s position in plane XY

\begin{equation} x = u_{x}t + \frac{1}{2}a_xt^2 \\ \end{equation}

Replacing

ux = 8.0 m/s

t = 4.0 second

ax = a Cos37° (Just X component of acceleration affects position of particle along X axis)

Value of Cos37° = 0.7986
a = 1.5 m/s2

ax = 1.5 × 0.7986 = 1.1979

ax = 1.1979 m/s2

Putting all these values into equation for finding x

\begin{equation} x = u_{x}t + \frac{1}{2}a_xt^2 \\ x = 8.0 × 4.0 + \frac{1}{2} × 1.1979 × (4.0)^2 \\ x = 32.0 + \frac{1}{2} × 1.1979 × 16.0 \\ x = 32.0 + \frac{1}{2} × 19.1664 \\ x = 32.0 + 9.6 \\ x = 41.6 \text{ m} \end{equation}

Therefore X position of particle at time t = 4.0 second will be 41.6 m

Similar to this let’s find out what’s Y position of particle at time t = 4.0 second

uy = 0 (Particle have initial velocity only along X axis)

t = 4.0 second

ay = a Sin37° (Value of acceleration only along Y axis matter to calculate particle’s velocity along Y axis)

ay = 1.5 × 0.60181 = 0.902715

ay = 0.902715 m/s2

Putting all these values in Second Equation of Motion

\begin{equation} y = u_yt + \frac{1}{2}a_yt^2 \\ y = 0 × 0 + \frac{1}{2} × 0.902715 × (4.0)^2 \\ y = \frac{1}{2} × 0.902715 × 16.0 \\ y = 7.22 \text{ m} \end{equation}

Therefore X position of particle at time t = 4.0 second will be 7.22 m

Therefore at time t = 4.0 second, particle is at (7.22, 41.6) point in XY plane.

Summarising the answers for this question
If a particle is moving at velocity 8.0 m/s in X-axis direction and it’s acceleration is 3.5 m/s2 at angle 37° with respect to X-axis then at time t = 4.0 second it’s velocity will be 13.3 m/s at angle 15.74° with respect to X-axis and it will be at point (7.22, 41.6) in XY plane.

Graph showing - If a particle is moving at velocity 8.0 m/s in X-axis direction and it's acceleration is 3.5 m/s2 at angle 37° with respect to X-axis then at time t = 4.0 second it's velocity will be 13.3 m/s at angle 15.74° with respect to X-axis and it will be at point (7.22, 41.6) in XY plane.

8. A ball is thrown from a field with a speed of 12.0 m/s at an angle of 45° with the horizontal. At what distance will it hit the field again? Take g = 10.0 m/s2

Horizontal Range of a projectile thrown at an angle θ with respect to X-axis will be u2Sin2θ/g
Where
u = Initial Velocity at which projectile is thrown

θ = Angle of initial velocity of projectile with X-axis

g = Acceleration due to gravity

Let’s draw this problem as a diagram for better understanding what’s actually happenning.

\begin{equation} \text{Horizontal Range} = \frac{u^2 Sin2θ}{g} \\ \end{equation}

Replacing
u = 12.0 m/s

θ = 45°

g = 10 m/s2

\begin{equation} \text{Horizontal Range} = \frac{u^2 Sin2θ}{g} \\ = \frac{(12.0)^2 × Sin(2 × 45°)}{10} \\ = \frac{144 × Sin90°}{10} \\ \text{Replacing Sin90° = 1} \\ = \frac{144 × 1}{10} = 14.4 m \end{equation}

Therefore it a projectile is thrown at velocity 12.0 m/s at angle of 45° with respect to horizontal axis then it’s horizontal range will be 14.4 m.


9. A man walks at a speed of 6 km/h for 1 km and 8 km/h for next 1 km. What is his average speed for the walk of 2 km?

The average speed of a particle in a time interval is defined as the distance travelled by the particle divided by the time interval. If particle travels a distance s in time t1 to t2, the average speed is defined as

\begin{equation} v_{av} = \frac {s} {t_{2} – t_{1}} \end{equation}

So average speed is just ratio of total distance covered to total time taken.

For 1 km walking at speed of 6 km/h
Distance = 1 km

Time = Distance/Speed = 1/6 h

For next 1 km walking at speed of 8 km/h
Distance = 1 km

Time = Distance/Speed = 1/8 h

So for entire journey of walking
Total Distane = 1 + 1 = 2 km

Total Time = 1/6 + 1/8 h

Average Speed = Total Distance/Total Time

\begin{equation} \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \\ = \frac{2}{\frac{1}{6} + \frac{1}{8}} \\ = \frac{2}{\frac{4 + 3}{24}} \\ = \frac{2}{\frac{7}{24}} \\ = \frac{2 × 24}{7} = \frac{48}{7} = 7 km/h \end{equation}

Therefore if a man is walking at 6 km/h for 1 km and then walks 1 km at 8 km/h speed. Then this for this 2 km of walking his average speed will be 7 km/h.


10. A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s.

Find

a) The Acceleration

b) Distance travelled during (t + 1)th second

Solution for part a)
The Acceleration

It’s given that particle is moving with constant acceleration a

Therefore

At time t second
a = 100/t

at = 100 (Equation 1)

At time (t + 1)th second
a = 150/(t + 1)

⇒ t + 1 = 150/a

a(t + 1) = 150 (Equation 2)

Subtracting (Equation 1) from (Equation 2)

a(t + 1) – at = 150 – 100

at + a – at = 50

a = 50 m/s2

Solution for part b)
Distance travelled during (t + 1)th second

For calculating distance travelled during (t + 1)th second
Let’s consider time interval from t second to (t + 1) second

For this time interval
u = 100 m/s

v = 150 m/s

a = 50 m/s2

Using Third Equation of Motion

\begin{equation} v^2 – u^2 = 2ax \\ (150)^2 – (100)^2 = 2 × 50 × x \\ 22500 – 10000 = 100x \\ 12500 = 100x \\ \frac{12500}{100} = x \\ 125 = x \end{equation}

Therefore Distance travelled during (t + 1)th second will be 125 metres.


11. A police inspector in a jeep is chasing a pickpocket on a straight road. The jeep is going at its maximum speed v (assumed uniform). The pickpocket rides on the motorcycle of a waiting friend when the jeep is at a distance d away, and the motorcycle starts with a constant acceleration a. Show that the pickpocket will be caught if v square root of (2ad).

Suppose the pickpocket motorcycle is caught by police car at time t after the motorcycle starts.

Distance travelled by motorcycle in this time interval t

S = at2/2

During same time interval distance travelled by police car

S + d = vt

So we have equations

S = at2/2 (For Motorcycle)


S + d = vt (For Police Car)

Replacing S with at2/2 in equation (For Police Car)

\begin{equation} \frac{at^2}{2} + d = vt \\ at^2 + 2d = 2vt \\ at^2 – 2vt + 2d = 0 \\ \text{This is just a Quadratic Equation in variable t} \\ \text{So solutions for this quadratic will be} \\ t = \frac{- (- 2v) ± \sqrt{(-2v)^2 – 4(a)(2d)}}{2a} \\ = \frac{2v ± \sqrt{4v^2 – 8ad}}{2a} \\ = \frac{2v ± \sqrt{4(v^2 – 2ad)}}{2a} \\ = \frac{2v ± 2\sqrt{v^2 – 2ad}}{2a} \\ = \frac{v ± \sqrt{v^2 – 2ad}}{a} \\ \implies t = \frac{v ± \sqrt{v^2 – 2ad}}{a} \\ \end{equation}

From this equation for value of t, it’s clear that in order for t to exist as a Real Number equation under square root should be positive or zero.

\begin{equation} v^2 – 2ad ≥ 0 \\ v^2 ≥ 2ad \\ v ≥ \sqrt{2ad} \end{equation}

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