Finding postion and velocity at t=0 given two different velocities and two different times?

An object moves with constant acceleration. At t= 2.50 s, the position of the object is x=2.00 m and its velocity is v= 4.50 m/s. At t= 7.00 s, v= -12.0 m/s. Find: (a) the position and the velocity at t= 0; (b) the average speed from 2.50s to 7.00 s, and (c) the average velocity from 2.50s to 7.00 s. I've tried using the kinematic equations of motion for constant acceleration but I'm not sure which time and velocity I'm supposed to use. please help thanks
Answers

Whome

Let's find that constant acceleration a = Δv / Δt a = (vf - vi) / (tf - ti) a = (-12.0 - 4.50) / (7.00 - 2.50) a = -3.67 m/s² (a) the position and the velocity at t= 0 v = v₀ + at 4.50 = v₀ + (-3.67)2.50 v₀ = 13.7 m/s s = s₀ + v₀(t) + ½at² 2.00 = s₀ + 13.7(2.50) + ½(-3.67)2.50² s₀ = -20.7 m we could have also used the other data set to find the velocity at t = 0 -12.0 = v₀ + (-3.67)7.00 v₀ = 13.7 m/s (b) the average speed from 2.50s to 7.00 s, average speed is total distance traveled over time. As the velocity changed direction, it must have been somewhere 0 m/s. v² = u² + 2as 0² = 4.50² + 2(-3.67)(x - 2.00) x = 4.76 we also need the position at t = 7.00 s = 2.00 + 4.50(7.00 - 2.50) + ½(-3.67)(7.00 - 2.50)² s = -14.875 sv(avg) = [(4.76 - 2.00) + (4.76 - (-14.875)) / (7.00 - 2.50) sv(avg) = 4.98 m/s (c) the average velocity from 2.50s to 7.00 s. average velocity is displacement over time v(avg) = (-14.875 - 2.00) / (7.00 - 2.50) v(avg) = -3.75 m/s If this is helpful to you, please vote a Best Answer.

az_lender

The acceleration is (-12.0 m/s - 4.50 m/s)/(4.50 s) = -3.6667 m/s^2. (a) So the velocity at t = 0 would have been 4.50 m/s - (-3.6667 m/s^2)(2.50 s) = 13.667 m/s. The average velocity in the first 2.50s was 9.0833 m/s, so the position at t = 0 was x = 2.00m - (9.0833 m/s)(2.50s) = -20.7 m. (c) (4.50 m/s - 12.0 m/s)/2 = -3.75 m/s. (b) For this, I'll take the average positive velocity, and the average negative velocity, and then do a weighted average of the two. The average positive v in the interval is 2.25 m/s, and the absolute value of the average negative v is 6.00 m/s; the time spent going in each direction is PROPORTIONAL to these averages, so the overall average speed is (2.25^2 + 6.00^2)/(2.25 + 6.00) = 4.98 m/s.