Question

The velocity of an object moving along a straight line is given by – v(t) = t 2 − 3t + 2 (a) Find the displacement of the object from t = 0 to t = 3. (b) Find the distance traveled from t = 0 to t = 3.

Answer 1

Answer:

a) Displacement of the object from t = 0 to t = 3 is 1.5 m

b)  Distance of the object from t = 0 to t = 3 is 1.83 m

Explanation:

Velocity, v(t) = t² - 3t + 2

a) Displacement is given by integral of v(t) from 0 to 3.

   $s=\int_{0}^{3}(t^2-3t+2)dt=\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_0^3=\frac{3^3}{3}-\frac{3^3}{2}+6=1.5m$

b) t² - 3t + 2 = (t-1)(t-2)

   Between 1 and 2,  t² - 3t + 2 is negative

   So we can write t² - 3t + 2 as -(t² - 3t + 2)

   Distance traveled

             $s=\int_{0}^{1}(t^2-3t+2)dt+\int_{1}^{2}-(t^2-3t+2)dt+\int_{2}^{3}(t^2-3t+2)dt\\\\s=\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_0^1-\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_1^2+\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_2^3\\\\s=\frac{1^3}{3}-\frac{3\times 1^2}{2}+2-\left ( \frac{2^3}{3}-\frac{3\times 2^2}{2}+4\right )+\frac{1^3}{3}-\frac{3\times 1^2}{2}+2+\frac{3^3}{3}-\frac{3\times 3^2}{2}+6-\left ( \frac{2^3}{3}-\frac{3\times 2^2}{2}+4\right )=1.83m$