Problem 5: Particle Motion


The position of a particle (in inches) moving along the x-axis after t seconds have elapsed is given by the following equation:

s = f(t) = t4 – 2t3 – 6t2 + 9t

(a) Calculate the velocity of the particle at time t.

(b) Compute the particle’s velocity at t = 1, 2, and 4 seconds.

(c) When is the particle at rest?

(d) When is the particle moving in the forward (positive) direction?

(e) Calculate total distance traveled by the particle (i.e., forwards and backwards) after t = 5 seconds.

(f) Calculate the acceleration of the particle after 4 seconds.

(g) When is the speed of the particle constant?


(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 true and to factor the equation. You will get the following:

(t – 1/2)(4t2 – 12t + 6)

Now, use the quadratic formula to solve the quadratic part, and you’ll see 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, 0.6339) and (2.366, ∞). Note that it doesn’t quite make sense to have negative time, so (0, 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, substitute 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. You find thats(0.6339) = 2.946152423, s(2.366) = –7.446152423, and s(5) = 270. Note that the negative x-intercept is ignored because you cannot move back in time.

These numbers represent how far the particle is from the origin at specific times. So, the particle 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 t = 2.366 seconds, the particle has traveled to the right 2.9 inches, back 2.9 inches to the origin, then left 7.4 more inches. It then moves 7.4 inches back to the origin and ends up 270 more inches to the right of it. The final answer is 290.785 inches.

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

(g) Set the acceleration equal to zero and solve using the quadratic equation: t = –0.618 sec or 1.618 sec.

Problem 4: B-B-B-Betty and the Jet (Related Rates)


During a taping for Circus of the Stars, beloved actress Betty White is shot out of a cannon. The firing goes completely awry and sends her on a collision course with a jet. As they converge, Betty and the jet plane at right angles to each other (see diagram below). Betty is 200 miles away from the point of impact and traveling at a constant rate of 600 mph. (Not even the laws of physics can slow Betty White!) The plane is 150 miles from impact and traveling at a constant rate of 450 mph.


At what rate is the distance d between Betty and the jet decreasing?


Consider the following diagram, which labels the legs of the right triangle as follows: b is the distance between Betty White and the point of impact and p is the distance between the plane and the point of impact.


The Pythagorean Theorem describes the relationship between the lengths of the sides of the triangle.

b2 + p2 = d2

Substitute b = 200 and p = 150 into the formula to solve for d, the distance between the two airborn objects at this moment.


You are asked to find the rate at which d decreases. In other words, you are calculating dd/dt. Apply implicit differentiation, with resepct to t.


Divide each of the terms by 2 and solve for dd/dt.


To calculate dd/dt, substitute all of the known information into the equation: b = 200, db/dt = –600, p = 150, dp/dt = –450, and d = 250. Note that db/dt and dp/dt are negative because the lengths of the legs of the right triangle are decreasing—the objects are on a collision course, so the distances between the objects and the point of impact are getting smaller.


The distance between Betty and the jet is decreasing at a rate of 750 mph.

Problem 3: Position and Velocity


An object is dropped from the second-highest floor of the Sears Tower, 1542 feet off of the ground. (The top floor was unavailable, occupied by crews taping for the new ABC special “Behind the Final Behind the Rose Final Special, the Most Dramatic Behind the Special Behind the Rose Ever.”)

(a) Construct the position and velocity equations for the object in terms of t, where t represents the number of seconds that have elapsed since the object was released.

(b) Calculate the average velocity of the object over the interval t = 2 and t = 3 seconds.

(c) Compute the velocity of the object 1, 2, and 3 seconds after it is released.

(d) How many seconds does it take the object to hit the ground? Report your answer accurate to the thousandths place.

(e) If the object were to hit a six-foot-tall man squarely on the top of the head as he (unluckily) passed beneath, how fast would the object be moving at the moment of impact? Report your answer accurate to the thousandths place.

Extra Credit: If the falling object killed the six-foot-tall man, is he actually luckier for not having to endure the Bachelor special taping on the top floor? (Spoiler alert: Yes.)


(a) The position function for a projectile is s(t) = –16t2 + v0t + h0, where v0  represents the initial velocity of the object (in this case 0) and h0 represents the initial height of the object (in this case 1,542 feet). Note that this position equation represents the height in feet of the object t seconds after it is released. Thus, the position equation is s(t) = –16t2 + 1,542. The vecocity equation v(t) is the derivative of the position equation: v(t) = –32t.

(b) Average velocity is the slope of the secant line, rather than the slope of the tangent line. Plug t = 2 and t = 3 into the position equation to calculate the height of the object at the boundaries of the indicated interval to generate two ordered pair: (2, 1478) and (3, 1398). Apply the slope formula from basic algebra to calculate the slope of the line passing through those points.


(c) Substitute t = 1, 2, and 3 into v(t).


(d) The object hits the ground when its position is s(t) = 0. Set the position equation equal to zero and solve for t.


(e) The problem asks you to calculate the velocity of the object when it is exactly six feet off of the ground, when s(t) = 6. Apply the same technique you completed in part (d), but instead of calculating the time t when the object’s position is 0, calculate the time t when its position is 6.


Now calculate the velocity of the object at that time: v(9.79795897113) = –32(9.79795897113) = –313.535 ft/sec.

Problem 2: Continuity


Let f(x) be the function defined below:


Determine whether f(x) is continuous at x = 0 and explain your answer.

Note: You may use a graphing calculator to examine the graph of f(x).


If f(x) if continuous at x = 0, its left- and right-hand limits exist at x = 0, and they are both equal to f(0). Consider the graph of the function below.

This sine curve is a “damped” function; it is already zoomed in quite far, but feel free to zoom in to your heart’s content. The function will wriggle its way to a height of 0 as you approach the y-axis from the right and from the left. Therefore, the general limit exists, and it is equal to 0.

According to the piecewise-defined function, f(0) = 0. (It’s a good thing, too, because substituting 0 intox2sin(1/x) would have been a deal-breaker. You’re not allowed to have a 0 in a denominator.)

Because the limit of f(x) exists as x approaches 0 and it equals f(0), you conclude that f(x) is continuous at 0.

Problem 1: Limits


Describe or draw a function, f(x), with the following characteristics:

  • f(x) has domain (–∞,8)
  • f(x) has range (–∞,9)
  • f(4) = 0; f(5) = 0; f(7) = 0
  • The limit, as x approaches –∞, of f(x) equals 9
  • The limit, as x approaches 8 from the left, of f(x) equals –∞
  • f(–1) = f(–3); f(–1) > f(–2)
  • f(1) = 1
  • The limit, as x approaches 1, of f(x) is 4

One Solution (others possible but similar):