A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t^2 – 5t)cm/sec^2. What will be the velocity of the particle?

Category: QuestionsA particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t^2 – 5t)cm/sec^2. What will be the velocity of the particle?
Editor">Editor Staff asked 11 months ago

A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t^2 – 5t)cm/sec^2. What will be the velocity of the particle?
 
(a) 27cm/sec
 
(b) 28 cm/sec
 
(c) 29 cm/sec
 
(d) 30 cm/sec
 
The question was posed to me in exam.
 
This question is from Linear First Order Differential Equations topic in portion Differential Equations of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

Correct choice is (c) 29 cm/sec
 
To elaborate: 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 = 3t^2 – 5t
 
Or, dv = 3t^2 dt – 5tdt
 
Or, ∫dv = 3∫t^2 dt – 5∫t dt
 
Or, v = t^3 – (5/2)t^2 + c  ……….(1)
 
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
 
Thus v = t^3 – (5/2)t^2 + 5
 
Or, dx/dt = t^3 – (5/2)t^2 + 5 ………..(2)
 
Thus, the velocity of the particle at the end of 4 seconds,
 
= [v]t = 4 = (4^3 – (5/2)4^2 + 5 ) cm/sec [putting t = 4 in (2)]
 
= 29 cm/sec