I'm having trouble with this problem.

Question

The question is: An object begins to move along the y axis and its position is given by the equation y = 5t2 − 3t − 2, with y in meters and t in seconds. (Express your answer in vector form.)

 

a.) What is the position of the object when it changes its direction?

 

b.) What is the object's velocity when it returns to its original position at
t = 0?
 

Answer 1

The equation you've been given is a quadratic equation, which means that the graph is a parabola. According to the problem, the parabola traces the height (along the y-axis) of the moving object.

 

Thinking about the U-shape of a parabola, you can imagine that the object starts out moving in one direction (in this case, downwards) as time goes by, then eventually changes direction to continue moving in the opposite direction (in this case, upwards).

 

So, where on the U-curve does the object change direction, from moving downwards to moving upwards? That's right at the parabola's vertex, its lowest point.

 

 

 

Part (a), therefore, just wants you to find the vertex of the parabola. You can do that by first finding the axis of symmetry, which gives you the x-coordinate of the vertex:

 

x = -b/(2a) = -(-3)/(2(5)) = 3/10

 

Now, plug that value for x into the equation to find the y-coordinate of the vertex:

 

y = 5(3/10)² - 3(3/10) - 2

y = 5(9/100) - 9/10 - 2

y = 9/20 - 18/20 - 40/20

y = -49/20

 

The position of the object when it changes its direction is -49/20 meters, after 3/10 seconds. In vector form, that point is (3/10, -49/20).

 

 

 

For (b), you'll first want to plug in t = 0 to find the objects original position:

 

y = 5(0)² - 3(0) - 2

y = 0 - 0 - 2

y = -2

 

So, to find out when the object returns to its initial position, plug in y = -2 and solve for t. You'll get two answers, one of which will be 0 and the other of which will mark the return of the object to its original position:

 

-2 = 5t² - 3t - 2

0 = 5t² - 3t

0 = t(5t - 3)

0 = t    OR    0 = 5t - 3

0 = t    OR    3 = 5t

0 = t    OR    3/5 = t

 

The object returns to its original position at t = 3/5 seconds.

 

For the object's velocity at t = 3/5 seconds, take the derivative of the original equation:

 

y' = 10t - 3

 

and plug in t = 3/5:

 

y' = 10(3/5) - 3

y' = 30/5 - 3

y' = 6 - 3

y' = 3

 

The value of the object's velocity when it returns to its original position is 3 m/s. The object is just moving along the y-axis, we're told, so it's moving at 3 m/s straight upwards. That vector is represented as (0, 3), since the object's x position isn't changing.

Answer 2

Assign values to t and compute the corresponding y result. Graph those 2 variables to establish the vector representation. Part 2 can be answered for the total time in t and distance traveled.