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[(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[(t^3)/(3)-8(t^2)/(2)+15t\right]^4_2`

`\\ \sf\longmapsto s=\left[(t^3)/(3)-4t^2+15t\right]^4_2`

`\\ \sf\longmapsto s=\left((4^3)/(3)-4(4)^2+15(4)\right)-\left((2^3)/(3)-4(2)^2+15(2)\right)`

`\\ \sf\longmapsto s=\left((64)/(3)-64+60\right)-\left((8)/(3)-16+30\right)`

`\\ \sf\longmapsto s=\left((64)/(3)-4\right)-\left((8)/(3)+14\right)`

`\\ \sf\longmapsto s=(64)/(3)-4-(8)/(3)-14`

`\\ \sf\longmapsto s=(64)/(3)-(8)/(3)-4-14`

`\\ \sf\longmapsto s=(46)/(3)-18`

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

• Take it +ve

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

`\\ \sf\longmapsto s=2.7km`