1.

A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t^2 – 4t + 5 cm/second. What will be the acceleration of the particle during first 3 seconds after the start?(a) 10cm/sec^2(b) 12cm/sec^2(c) 14cm/sec^2(d) 16cm/sec^2This question was posed to me in unit test.Question is from Calculus Application topic in division Application of Calculus of Mathematics – Class 12

Answer»

The CORRECT CHOICE is (c) 14cm/sec^2

Explanation: Let f be the acceleration of the particle in time t seconds. Then,

f = dv/DT = d(3t^2 – 4t + 5)/dt

= 6t – 4……….(1)

Therefore, the acceleration of the particle at the end of 3 seconds,

= [f]t = 3 = (6*3 – 4) cm/sec^2

= 14cm/sec^2



Discussion

No Comment Found

Related InterviewSolutions