Question

A particle moves along the x-axis. The velocity of the particle at time t is −2t+8 . What is the total distance traveled by the particle from t=1 to t=6?

Answer 1

Answer:

13

Step-by-step explanation:

The velocity is given by:

$\displaystyle v(t)= -2t + 8$

And the total distance given the velocity is given by:

$\displaystyle \int_{a}^{b}|v(t)|\, dt$

So, the total distance traveled by the particle from t = 1 to t = 6 will be:

$\displaystyle d=\int_{1}^{6}|-2t+8|\, dt$

The absolute value tells us that:

$\displaystyle |-2t+8| = \left\{ \begin{array}{ll} -2t+8 & \quad x \leq 4 \\ -(-2t+8) & \quad x > 4\\ \end{array} \right$

So, split the integral into two parts:

$\displaystyle d=\int_{1}^{4} (-2t+8)\, dt+\int_{4}^{6}(-(-2t+8))\, dt$

Integrate:

$\displaystyle d=(-t^2+8t)\Big|_{1}^{4}+(t^2-8t)\Big|_{4}^{6}$

Evaluate:

$\begin{aligned} d&=[(-4^2+8(4))-(-1^2+8)]+[(6^2-8(6))-(4^2-8(4)]\\ &= [16-7]+[-12-(-16)]\\ &=9+4 \\ &=13 \end{aligned}$