Question

a particle starts from rest at a point a and moves along a straight line ab with an acceleration 8-2t² m/s² at time t seconds. find the greatest speed of the particle in direction ab and distance covered by the particle in first two seconds of it's motion.​

Answer 1

Answer:

for velocity to be maximum.. its derivative should be 0

so, dv/dt = a = 0

=> 8-2t² = 0

=> t² = 4

or t = ±2

now, the maximum value of velocity is

v = integration of a = 8t-2t³/3 = 8(2)-2(2)³/3 = 16-16/3 = 32/3 m/s

and s = integration of V = 4t²-t⁴/6

so, s ( at t = 2) = 4(2)²-(2)⁴/6 = 16-16/6 = 80/6 = 40/3 m

Explanation:

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