1.

A particle is moving along the straight line OX and its distance x is in metres from O after t seconds from start is given by x = t^3 – t^2 – 5t. What will be the acceleration of the particle when it is at a distance 28 metres from O?(a) 20 m/sec^2(b) 22 m/sec^2(c) 24 m/sec^2(d) 26 m/sec^2The question was asked by my college professor while I was bunking the class.My question comes from Calculus Application in portion Application of Calculus of Mathematics – Class 12

Answer»

Right option is (b) 22 m/sec^2

Best explanation: We have, x = t^3 – t^2 – 5t……….(1)

When x = 28, then from (1) we get,

t^3 – t^2 – 5t = 28

Or t^3 – t^2 – 5t – 28 = 0

Or (t – 4)(t^2 + 3t +7) = 0

Thus, t = 4

Let v and f be the velocity and ACCELERATION respectively of the particle at time t seconds. Then,

v = dx/DT = d(t^3 – t^2 – 5t)/dt

= 3t^2 – 2T – 5

And f = dv/dt = d(3t^2 – 2t – 5)/dt

= 6t – 2

Therefore, the acceleration of the particle at the end of 4 seconds i.e., when the particle is at a distance of 28 METRES from O,

[f]t = 4 = (6*4 – 2) m/sec^2

= 22 m/sec^2


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