Video: APCALC02AB-P1A-Q14-157195614187

A particle moves along the π‘₯-axis. The velocity of the particle at time 𝑑 is 3𝑑² + 5𝑑. What is the total distance travelled by the particle from time 𝑑 = 0 to time 𝑑 = 4?

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Video Transcript

A particle moves along the π‘₯-axis. The velocity of the particle at time 𝑑 is three 𝑑 squared plus five 𝑑. What is the total distance travelled by the particle from time 𝑑 equals zero to time 𝑑 equals four?

We’re told that the particle moves along the π‘₯-axis. This is a single straight line. So we can use the rules for rectilinear motion. Let’s begin by recalling the relationship between velocity and displacement when considering rectilinear motion.

Now I said displacement as velocity is a vector quantity, as is displacement. Distance is a scalar quantity. So we’ll soon use the fact that it’s the magnitude of the displacement. Velocity is the rate of change of displacement over time. In other words, if 𝑠 is a function for displacement in time, then 𝑣, velocity, is equal to d𝑠 by d𝑑. It’s the derivative of the function 𝑠 for displacement with respect to time.

We can reverse this idea and use the fact that integration is the reverse process for differentiation. We can say that 𝑠 is equal to the integral of the function 𝑣 with respect to 𝑑. To find the displacement of the particle between zero and four seconds then, we integrate our expression for velocity with respect to time between the limits of zero and four.

Remember, when we integrate, we add one to the power and then divide by that new number. So the integral of three 𝑑 squared is three 𝑑 cubed divided by three, which is just 𝑑 cubed. And the integral of five 𝑑 is five 𝑑 squared divided by two. And of course, we’re going to evaluate this between zero and four. That’s four cubed plus five times four squared over two minus zero cubed plus five times zero squared over two. Zero cubed is zero, and five times zero squared divided by two is zero. So we get 64 plus five times 16 over two. We divide through by two here. And we see that the displacement becomes 64 plus 40, which is 104 or 104 units.

Remember, we’re trying to find the distance travelled during this time. And distance is equal to the magnitude of the displacement. Well, the magnitude of 104 is just 104. So the total distance travelled by the particle from time 𝑑 equals zero to time 𝑑 equals four is 104 or 104 units.

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