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**Introductory Exercise 3.2**

**Ques 1:**** A ball is thrown vertically upwards. Which quantity remains constant among, speed, kinetic energy, velocity and acceleration?****Ans: **acceleration**Sol: **Acceleration (due to gravity).**Ques 2: Equation does not seem dimensionally correct, why?****Ans: ****Sol: **is physically correct as itgives the displacement of the particle in t^{th} second (or any time unit).

s_{t} = Displacement in t seconds - displacement in (t - 1) seconds

Therefore, the given equation is dimensionally incorrect.**Ques 3: Can the speed of a particle increase as its acceleration decreases? If yes give an example.****Ans: **Yes, in simple harmonic motion**Sol: **Yes. When a particle executing simple harmonic motion returns from maximum amplitude position to its mean position the value of its acceleration decreases while speed increases.**Ques 4: The velocity of a particle moving in a straight line is directly proportional to 3/4th power of time elapsed. How does its displacement and acceleration depend on time?****Ans: t ^{7/4}, t^{-1/4}**

…(i)

∴

or

i.e., s ∝ t

Differentiating Eq. (i) w.r.t. time t,

⇒ a ∝ t

= 240 - 180

= 60 m (in the upward direction)

Distance covered (D) by the particle

Time to attain maximum height

It implies that particle has come back after attaining maximum height (h) given by

∴ D = 80 + (80 - 60)

= 100 m

∴

or dx = (40 - 10t) dt

or x = ∫ (40 - 10t) dt

or x = 40t - 5t

As at t = 0 the value of x is zero.

c = 0

∴ x = 40t - 5t

For x to be 60 m.

60 = 40t - 5t

or t

∴ t = 2 s or 6 s

**Sol: ****Ques 8: A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is v _{1} and at time t = t is v_{2}. The average velocity of the particle in this time interval is Is this statement true or false?**

∴ at = v

⇒ t = 25 s

Average velocity = 125 m/5 s (downwards)

= 25 m/s (downwards)

(b) 43 .5 m

∴

At t = 2 s

a = 5 - 2 x 2

= 1 m/s

From Eq. (i),

∴ x = ∫ (10 + 5t - t^{2}) dt

or

As, at t = 0 the value of x is zero

c = 0

∴

Thus, at t = 3 s

= 30 + 22.5 - 9

= 43.5 m**Ques 11: Velocity of a particle at time t = 0 is 2 m/s. A constant acceleration of 2 m/s ^{2} acts on the particle for 2 seconds at an angle of 60° with its initial velocity. Find the magnitude of velocity and displacement of particle at the end of t = 2 s.**

= 4√3 m

…(i)

From Eq. (i),

∴

Taking initial displacement to be zero.

Yes. As explained below.

implies that initial velocity of the particle is and the acceleration is

∴

∴

or, x

(The above is the equation to trajectory) x = 2t

∴

y = t

∴

Thus,

__Introductory Exercise__** 3.3**

**Ques 1: Figure shows the displacement-time graph of a particle moving in a straight line. Find the signs of velocity and acceleration of particle at time t = t _{1} and t = t_{2}.**

v = tan θ

Corresponding v-t graph will be

Acceleration at t = t_{1} :

As α < 90°, a t_{1} is + ive constant.

Acceleration at t = t_{2}

**Ques 2: A particle of mass m is released from a certain height h with zero initial velocity. It strikes the ground elastically (direction of its velocity is reversed but magnitude remains the same). Plot the graph between its kinetic energy and time till it returns to its initial position.****Ans:****Sol: **Let the particle strike ground at time t velocity of particle when it touches ground

would be gt. KE of particle will be i.e., KE ∝ t^{2}. While going up the velocity will get - ive but the KE will remain. KE will reduce to zero at time 2 t when the particle reaches its initial position.**Ques 3: A ball is dropped from a height of 80 m on a floor. At each collision, the ball loses half of its speed. Plot the speed-time graph and velocity-time graph of its motion till two collisions with the floor. [Take g = 10 m/s ^{2}]**

Time taken to reach ground

Speed of ball (just after first collision with floor)

= 40/2 = 20 m/s

Time to attain maximum height

∴ Time for the return journey to floor = 2 s.

Corresponding velocity-time will be

⇒ h = 2 (t - 2)

Particle will attain its initial velocity i.e., net increase in velocity of the particle will be zero when,

area under a-t graph = 0

or 3 - (t - 2)

or (t - 2)

or t - 2 = ± √3

or t = 2 ± √3

At time t = 2 + √3 s

**Introductory Exercise 3.4**

**Ques 1: Two balls A and B are projected vertically upwards with different velocities. What is the relative acceleration between them?Ans:** zero

α

∴ Relative displacement (i.e., distance between A and B) would be

or s = (v

(b) 80 m

v = speed of river flow

Time to cross river**Ques 4: An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. Wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. ****(a) Find the direction in which the pilot should head the plane to reach the point B. ****(b) Find the time taken by the plane to go from A to B.****Ans: **

(b) 50 min**Sol: **Let C be the point along which pilot should head the plane.

Apply sine formula in Δ ABC

⇒

= 2989 s

= 50 min**Ques 5:**** Two particles A and B start moving simultaneously along the line joining them in the same direction with acceleration of 1 m/s ^{2} and 2 m/s^{2} and speeds 3 m/s and 1 m/s respectively. Initially A is 10 m behind B. What is the minimum distance between them?**

v

Acceleration of A w.r.t. B = 1 - 2 = - 1 m/s

Velocity of A w.r.t. B = 3 - 1 = 2 m/s

Initial displacement of A w.r.t. B = - 10 m

At time relative displacement of A w.r.t. B

or s = - 10 + 2t - 0.5t

For s to be minimum

or 2 - (0.5 x 2t) = 0

i.e., t = 2 s

∴ s

= - 10 + 4 - 2

= - 8 m

Minimum distance between A and B = 8 m.

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