1.

A particle is moving in a straight line is at a distance x from a fixed point in the straight line at time t seconds, where x = 2t^3 – 12t + 11. What is the average acceleration of the particle at the end of 3 seconds?(a) 28 cm/sec^2(b) 30 cm/sec^2(c) 32 cm/sec^2(d) 26 cm/sec^2This question was addressed to me in homework.I would like to ask this question from Calculus Application in portion Application of Calculus of Mathematics – Class 12

Answer»

The correct CHOICE is (b) 30 cm/sec^2

The explanation is: We have, x = 2t^3 – 12t + 11……….(1)

Let v and f be the velocity and acceleration respectively of the particle at time t seconds.

Then, v = dx/dt = d(2t^3 – 12t + 11)/dt

= 6t^2 – 12……….(2)

And f = dv/dt = d(6t^2 – 12)/dt

= 12t……….(3)

Putting the value of t = 2 in (3),

THEREFORE, the acceleration of the particle at the end of 2 seconds,

12t = 12(2)

= 24 cm/sec^2

Now putting the value of t = 2 in (2),

We get the displacement of the particle at the end of 2 seconds,

6t^2 – 12 = 6(2)^2 – 12

= 12 cm/sec……….(4)

And putting the value of t = 3 in (2),

We get the displacement of the particle at the end of 3 seconds,

6t^2 – 12 = 6(3)^2 – 12

= 42 cm/sec……….(5)

Thus, change in velocity is, (5) – (4),

=42 – 12

= 30cm/sec.

Thus, the average acceleration of the particle at the end of 3 seconds is,

= (change of velocity)/time

= (30 cm/sec)/1 sec

= 30 cm/sec^2


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