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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,

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