Problem of the Week #6

Problem 6: 1998-1999

The position of a particle (in inches) moving along the x-axis is given by the equation:

s = f(t) = t⁴ - 2t³ - 6t² + 9t

(a) Find velocity at time t.

(b) What is the velocity after 1, 2, and 4 seconds?

(d) When is it moving forward?

(e) What is the total distance traveled (i.e. forwards and backwards) in 5 seconds?

(f) What is the acceleration at t seconds? at 4 seconds?

(g) When is the speed constant (acceleration = 0)?

Solution: (a) The velocity is the derivative of position, so the velocity is v(t) = 4t³- 6t²- 12t + 9.

(b) Simply plug into the velocity equation to get: v(1)=-5 in/sec, v(2)=-7 in/sec, v(4)=121 in/sec.

(c) If you graph the velocity function on your calculator, you see that it appears to pass through x = -1.5. Use synthetic division to ensure that this is the truth and to factor the equation. You will get:

. Now, use the quadratic formula to solve the quadratic part, and you'll get that the velocity equals zero (in other words, is stopped) when t = -1.5, 0.6339745962, 2.366025404. Even though you can round to the third decimal place, you need to use these values for the remainder of the problem.

(d) If you plug in values into the velocity equation between the x-intercepts above, you will get positive values on the intervals (-1.5, .6339) and (2.366, infinity). Note that it doesn't quite make sense to have nagative time, so (0, .6339) is just as acceptable, and perhaps more so, for the first interval. We do this because positive velocity implies forward movement.

(e) First, plug the "turn points" you found in part (c) into the position equation. When the velocity equals zero in this problem, the particle is stopping because it is turning to go the other way. We find that s(.6339)=2.946152423, s(2.366)=-7.446152423, s(5)=270. Note that the negative x-intercept is ignored because we cannot move back in time. These numbers represent how far the particle is away from the origin at the specific times. So, the particles moves 2.9 inches to the right of the origin, then moves 7.44 inches left of it, and finally ends up 270 inches to the right of it. By the time I have reached 2.366 seconds, then, I have traveled to the right 2.9 inches, back 2.9 inches to the origin, then left 7.4 more inches. I then move 7.4 inches back to the origin and end up 270 more inches to the right of it. Don't forget to count in the doubling back when you calculate this answer, which is 290.785 inches.

(f) The acceleration is the derivative of velocity, so a(t) = v'(t) = 12t² - 12t - 12. The acceleration at 4 seconds is a(4) = 132 in/sec².

(g) We need to set the acceleration = 0 to solve this, so use the quadratic formula to get that speed is constant when t = -0.618 sec or 1.618 seconds.

Superbowl of Calculus