Question

The velocity of a particle is dependent on the time as v=k(t-1) where k = 2 m/s2 the distance covered in first three seconds will be
(a) 18 m (b) 5 m (C) 3 m (d) 6 m


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Answer 1

this is ur solution

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Answer 2

Finding v = 0

v = 0 → k(t - 1) = 0

→ t = 1

So at t = 1 velocity becomes zero, that means it will take a U-turn and hence, direct integrating the given function will yield displacement and not distance.

So, to find distance, first of all integrate from 0 to 1 second, and then integrate separately from 1 to 3 second.

Let the distance be denoted by s

→ ds/dt = v

→ ds = vdt

→ ds = k(t - 1)dt

After integrating,

s = k(t²/2 - t)

Put value of k = 2 and put the limits from t = 0 to t = 1

→ 2(1/2 - 1) = -1

This is the displacement in negative direction. So distance = 1 m (from 0 to 1 second)

Now, put the limits from t = 1 to t = 3

s = 2(3²/2 - 3) - 2(1/2 - 1)

s = 2(9/2 - 3) - (-1)

s = 2(3/2) + 1

s = 4

So total distance would be, 4 + 1 = 5 m