v = t^{2} – t + 6

- Find the displacement of the particle during the time period 1 ≤ t ≤ 4.
- Find the distance travelled during this time period.

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Solution:

Given,

v = t^{2} – t + 6

**a) **

As we know,

v = ds / dt

ds = v . dt

Integrating both side,

\(\int ds = \int v \enspace dt\)

Now, We will integrate in the interval [1, 4]

\(\int_{1}^{4} ds = \int_{1}^{4} v \enspace dt\)

\( = \int_{1}^{4} (t^2 – t – 6) \enspace dt\)

\( = \left [ \frac{t^3}{3} – \frac{t^2}{2} – 6t \right ]_{1}^{4}\)

\(= \left [ \frac{64}{3} – 8 – 24 \right ] – \left [ \frac{1}{3} – \frac{1}{2} – 6 \right ]\)

= – 9 / 2

Therefore, displacement travelled by particle is -9/2 m

**b)**

Since, distance travelled so, V = 0

then, t^{2} – t + 6 = 0

or, (t – 3)(t + 2) = 0

or, t = 3, -2

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## Discussion