A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the velocity of the particle?a)27cm/secb)28 cm/secc)29 cm/secd)30 cm/secCorrect answer is option 'C'. Can you explain this answer?

Question

A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 &ndash; 5t)cm/sec2. What will be the velocity of the particle?a)27cm/secb)28 cm/secc)29 cm/secd)30 cm/secCorrect answer is option 'C'. Can you explain this answer?

Answer

Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t² – 5t
Or, dv = 3t² dt – 5tdt
Or, ∫dv = 3∫t² dt – 5∫t dt
Or, v = t³ – (5/2)t² + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t³ – (5/2)t² + 5
Or, dx/dt = t³ – (5/2)t² + 5 ………..(2)
Thus, the velocity of the particle at the end of 4 seconds,
= [v]_{t = 4} = (4³ – (5/2)4² + 5 ) cm/sec [putting t = 4 in (2)]
= 29 cm/sec

Answer

To find the equation of motion for the particle, we need to integrate the acceleration function.

Given:
Velocity (v) = 5 cm/sec
Acceleration (a) = 3t^2 cm/sec^2

Integrating acceleration gives us velocity:
v = ∫(3t^2) dt
v = t^3 + C1

Given that the initial velocity is 5 cm/sec, we can substitute this value into the equation:
5 = (0^3) + C1
C1 = 5

So, the equation for velocity is:
v = t^3 + 5

Now, to find the equation for displacement, we need to integrate velocity:
s = ∫(t^3 + 5) dt
s = (1/4)t^4 + 5t + C2

Since the particle starts from the origin (s = 0 at t = 0), we can substitute these values into the equation:
0 = (1/4)(0^4) + 5(0) + C2
C2 = 0

Therefore, the equation for displacement is:
s = (1/4)t^4 + 5t

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