Question
A race car moves such that its position fits
the relationship x=(5m/s) t + (.75m/s3 ) t3 where x is measured in meters and t
in seconds. A)Plot a graph of position versus time. B)Determine the
instantaneous velocity at t=4s using time intervals of .4s, .2s, and .1s.
I'm not looking for any answers, just a way to start B. For some reason, it's not clicking and it's not hard at all. Thanks ahead of time.
the relationship x=(5m/s) t + (.75m/s3 ) t3 where x is measured in meters and t
in seconds. A)Plot a graph of position versus time. B)Determine the
instantaneous velocity at t=4s using time intervals of .4s, .2s, and .1s.
I'm not looking for any answers, just a way to start B. For some reason, it's not clicking and it's not hard at all. Thanks ahead of time.
Answers
B) The velocity at time t is the derivative of the x(t) function. Therefore
V(t) = 5 + 2.25 t^2. You can use that formula for the exact instantaneous velocity. At t=4 s, the instantaneous velocity is 5 + 2.25*16 = 41 m/s.
They are NOT asking you to use differential calculus, but instead to compute the position at x= t and t + delta t, and divide by the interval delta t. For delta t = 0.1 s, you get
V = [x(4 +.1) - x(4)]/(0.1 s)
= (72.19 - 68.00)/0.1 = 41.9 m/s
V(t) = 5 + 2.25 t^2. You can use that formula for the exact instantaneous velocity. At t=4 s, the instantaneous velocity is 5 + 2.25*16 = 41 m/s.
They are NOT asking you to use differential calculus, but instead to compute the position at x= t and t + delta t, and divide by the interval delta t. For delta t = 0.1 s, you get
V = [x(4 +.1) - x(4)]/(0.1 s)
= (72.19 - 68.00)/0.1 = 41.9 m/s
The velocity will be the slope of the posistion-time graph. Draw a tangent to the curve at the points desidred, and draw a triangle , then determine the slope (delta position/deltatime).