Expert: Ahmed Salami - 8/29/2009

Question
I have a question saying a car is at position x= the square root of(7t+2).  I need to find the instantaneous velocity at 3 seconds.  They want me to get closer and closer like from 3 to 3.1, 3 to 3.01, 3 to 3.001, and so on.  What equation should I use for this?

Answer
Hi Brandon,
The position of the car x at any time t is represented by the relation
x = √(7t+2)
Lets say we want to find the average velocity of the car between the times t = 3 and 3.1, we know that velocity is change in displacement divided by change in time and so we have
v = {√[7(3.1) + 2] - √[7(3) + 2]} / (3.1 - 3)
 = (4.8683 - 4.7958)/(0.1)
 = 0.0725/0.1
 = 0.725
Between t = 3 and 3.01
v = {√[7(3.01) + 2] - √[7(3) + 2]} / (3.01 - 3)
 = (4.803124 - 4.795832)/(0.01)
 = 0.007292/0.01
 = 0.7292
Between t = 3 and 3.001
v = {√[7(3.001) + 2] - √[7(3) + 2]} / (3.001 - 3)
 = (4.7965613 - 4.7958315)/(0.001)
 = 0.007298/0.001
 = 0.7298
Continuing in this manner for values closer to 3 such as 3.0001 and 3.00001 we can get as close as we want to 3 but we would realise that the value of the velocity reaches a limit of 0.7298 to four decimal places which is the instantaneous velocity at t = 3. Theoretically, the instantaneous velocity at t = 3 is the average velocity between t = 3 and, say, t = 3+ (i.e a number very very close to 3 almost coinciding with but not equal to 3, you get the idea).
The true value can be found using the methods of calculus.
v = dx/dt
 = 7/[2√(7t+2)]
and at t = 3
v = 0.7298

The lateness of the solution is regretted.

Regards