Question

suppose an object traveling in a straight line has a velocity function given by v(t)= t^2 -8t+ 15 km/hr. Find the displacement and distance traveled by the object from t=2 to t=4 hours.

Answer 1

v=t^2-8t+15

• It has upper limit 4 and lower limit 2

$\boxed{\sf {\displaystyle{\int}^b_a}x^ndx=\left[\dfrac{x^{n+1}}{n+1}\right]^b_a}$

$\\ \sf\longmapsto s={\displaystyle{\int}}vdt$

$\\ \sf\longmapsto s={\displaystyle{\int^4_2}}t^2-8t+15$

$\\ \sf\longmapsto s=\left[\dfrac{t^3}{3}-8\dfrac{t^2}{2}+15t\right]^4_2$

$\\ \sf\longmapsto s=\left[\dfrac{t^3}{3}-4t^2+15t\right]^4_2$

$\\ \sf\longmapsto s=\left(\dfrac{4^3}{3}-4(4)^2+15(4)\right)-\left(\dfrac{2^3}{3}-4(2)^2+15(2)\right)$

$\\ \sf\longmapsto s=\left(\dfrac{64}{3}-64+60\right)-\left(\dfrac{8}{3}-16+30\right)$

$\\ \sf\longmapsto s=\left(\dfrac{64}{3}-4\right)-\left(\dfrac{8}{3}+14\right)$

$\\ \sf\longmapsto s=\dfrac{64}{3}-4-\dfrac{8}{3}-14$

$\\ \sf\longmapsto s=\dfrac{64}{3}-\dfrac{8}{3}-4-14$

$\\ \sf\longmapsto s=\dfrac{46}{3}-18$

$\\ \sf\longmapsto s=15.3-18$

• Take it +ve

$\\ \sf\longmapsto s=|-2.7|$

$\\ \sf\longmapsto s=2.7km$