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

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.