Question

The position of a body moving along x axis is x=8.5+2.5t^2.Find the velocity during 2s-4s

Answer 1

Since the displacement x depends on time, the velocity will be the first derivative of this displacement wrt time. Thus,

v = d (8.5 + 2.5t²) / dt

v = 0 + 2.5(2t)

v = 5t

For t1 = 2s,

v1 = 5 × 2 = 10 m s^(-1)

And for t2 = 4s,

v2 = 5 × 4 = 20 m s^(-1)

Now, for the average velocity between these two times, we have to find the rate of change of displacement during this time interval, i.e.,

v = (x2 - x1) / (t2 - t1)

Remember, mean velocity is not meant here.

So, for time t1,

x1 = 8.5 + 2.5(t1)²

and for time t2,

x2 = 8.5 + 2.5(t2)²

Then,

v = (x2 - x1) / (t2 - t1)

v = ((8.5 + 2.5(t2)²) - (8.5 + 2.5(t1)²)) / (t2 - t1)

v = (8.5 + 2.5(t2)² - 8.5 - 2.5(t1)²) / (t2 - t1)

v = 2.5[(t2)² - (t1)²] / (t2 - t1)

v = 2.5(t2 - t1)(t1 + t2) / (t2 - t1)

v = 2.5(t1 + t2)

Taking t1 = 2s and t2 = 4s,

v = 2.5(2 + 4)

v = 2.5 × 6

v = 15 m s^(-1)

Hence this is the average velocity. Here it is equal to the mean velocity.

v = (v1 + v2) / 2

v = (10 + 20) / 2

v = 15 m s^(-1)

This is because the acceleration is constant in this motion and is not dependent on time. Else both mean and average velocity may vary each other.

Here, acceleration, a = 5 m s^(-2).

Acceleration is found as the first derivative of velocity wrt time.

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