An object moves with constant velocity from (4, −8) to (−5, 4), taking three seconds for the trip. Where is the object four seconds after it leaves (4, −8)?

Question

mathmale · Answer

Have you studied "vectors" yet, or are you to do this problem without vectors?

To calculate the velocity of this object as it travels from the 1st point to the 2nd, you need to find the distance between the 2 given points and then divide that distance by 3 sec.

calculusxy · Answer

I thought about making parametric for this (where t is the time every 3 seconds)
x = 4 - 9t
y = -8 + 12t

calculusxy · Answer

Can I divide those times by 3 to find the movement every 1 second. 
x = 4 - 3t
y = -8 + 4t 

Then I would get (if t = 4) 
x = 4 - 3(4) = -8
y = -8 + 4(4) = 8 

(8, -8) ?

mathmale · Answer

If you mult. this velocity by 3 sec, you'll arrive at the distance between the 2 points.
If you mult. this velocity by 4 sec, you'll be calculating the distance the object has traveled after 4 sec have elapsed since it left the 1st point.

mathmale · Answer

There are several ways in which you could approach this problem.  Vectors are one approach; parametric equations are another.  I'd say:  use the simplest approach possible, and then, if you have the time and the interest, try a different approach and compare your results.

sshayer · Answer

|dw:1481597503608:dw|

mathmale · Answer

With the method I'm suggesting, you will still have to find the coordinates of the final destination of the object after 4 sec.

calculusxy · Answer

Thank you! Can you help me with another question please?

sshayer · Answer

why, am i wrong?

sshayer · Answer

How can x be positive?

sshayer · Answer

x ,changes from 4 to -5 in three seconds in the next second negative value has to increase as motion is in a straight line.

sshayer · Answer

correction,y increases from -8 to 4 in 3 seconds and in next second it will increase further.