The position of a particle as a function of time is given by x = (1.8 m/s) t +(-

Question

The position of a particle as a function of time is given byx = (1.8 m/s)t +(-2.7m/s2)t2.

(b) Find the average velocity of the particle from t =0.45 s to t = 0.55 s.
2 ? m/s
(c) Find the average velocity from t = 0.49 s tot = 0.51 s.
3 ? m/s Please include as much detail as possible! Thankyou! The position of a particle as a function of time is given byx = (1.8 m/s)t +(-2.7m/s2)t2.

(b) Find the average velocity of the particle from t =0.45 s to t = 0.55 s.
2 ? m/s
(c) Find the average velocity from t = 0.49 s tot = 0.51 s.
3 ? m/s Please include as much detail as possible! Thankyou!

Explanation / Answer

You are given the position of a particle as a function oftime. To find the velocity of the particle as a function of time,you must compute the derivative of the postion function. To do that, use the power rule: where f '(x) =n(t)n-1 So in this case, the the velocity of the particle as afunction of time should be: 1.8 - (2.7)2(t)2-1 which equals: v(t) = 1.8-5.4t Now for question b and c, u can simply plug in your two "t"values and find the change in velocity.
PS: Sorry if I posted this more than once, I'mexperience major lag right now.

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