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 distance from the origin at the end of 4 seconds?a)30(4/3)b)30(2/3)c)30d)UnpredictableCorrect answer is option 'B'. 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 distance from the origin at the end of 4 seconds?a)30(4/3)b)30(2/3)c)30d)UnpredictableCorrect answer is option 'B'. Can you explain this answer?

Answer

The given acceleration at time t seconds is (3t^2).

To find the particle's velocity at time t seconds, we can integrate the acceleration function with respect to time.

∫(3t^2) dt = t^3 + C

Since the particle starts from the origin with a velocity of 5 cm/sec, we can find the constant C by substituting t = 0 into the velocity equation.

0^3 + C = 5
C = 5

Therefore, the velocity function is given by:

v(t) = t^3 + 5 cm/sec

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 – 5t dt
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)
Or, dx = t³ dt – (5/2)t² dt + 5 dt
Integrating this we get,
x = (1/4)t⁴ – (5/2)t³/3 + 5t + k ……….(3)
By the problem, x = 0, when t = 0; hence, from (3) we get, k = 0.
Thus, x = (1/4)t⁴ – (5/6)t³ + 5t ……….(4)
Thus, the velocity of the particle at the end of 4 seconds,
= [x]_{t = 4} = (1/4)4⁴ – (5/6)4³ + 5(4) [putting t = 4 in (4)]
= 30(2/3) cm

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