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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?

User Fody
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1 Answer

10 votes

Answer:

13

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}

User Seyi Aka Architect
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