Answer:
4.408 m/s, 4.102 m/s, 4.026 m/s
Explanation:
The question is incomplete. The text of the original question states:
A race car moves such that its position fits the relationship
:
[tex]x=(4.0 m/s)t + (0.85 m/s^3) t^3[/tex]
where x is measured in meters and t in seconds. Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s.
We can find the instantanoues velocity of the car at any time t by calculating the derivative of the position, so we find:
[tex]v(t) = x'(t) = 4.0 m/s + 3\cdot (0.85 m/s^2) t^2 = 4.0 m/s + (2.55 m/s^2) t^2[/tex]
And now we just need to substitute t=0.40 s, 0.20 s, and 0.10 s to find the corresponding instantaneous velocity:
[tex]v(0.40) = 4.0 + 2.55 (0.40)^2 = 4.408 m/s\\v(0.20) = 4.0 + 2.55 (0.20)^2 = 4.102 m/s\\v(0.10) = 4.0 + 2.55 (0.10)^2 = 4.026 m/s[/tex]
