Physics, acceleration

Question

1)

A particle moves with a constant acceleration. It's final velocity is 7m/s and its acceleration is -3m/s^2. Find the initial velocity if the particle travels a distance of 7m

2)

A particle moves with a constant acceleration. It's original velocity is 2m/s and it's final velocity is -5m/s. Find the time takes to cover the distance -2m

3)

A particle moves with a constant acceleration 3m/s^2, it's original velocity is 3m/s. Find the distance in meters the particle has travelled after 6s

Answer 1

The series of kinematic equations of motion includes:

1) x_f = x_i + ½(v_f + v_i)t

2) v_f = v_i + at

3) x_f = x_i + v_it + ½ at²

4) v_f² = v_i² + 2a(x_f − x_i)

It's just a matter of trying to figure out which one to use based on the problem you have.

For problem 1), the elements "in play" are final velocity (v_f), acceleration (a), initial velocity (v_i) and distance (x). So we pick equation #4 because it has those same elements in it. Since we are looking for initial velocity, we tweak the equation a little to solve for vI before we get started. Also, notice that we can treat x_f - x_i as the total distance traveled. So v_i² = v_f² - 2a(x_f - x_i) or v_i = square root(v_f² - 2a(x_f - x_i)). Now we just plug in the numbers to get v_i = square root(7² - 2(-3)(7)) = square root(49 + 42) = √91 = 9.54 m/s. However, there is only one significant figure in the initial givens, so we round to 1 sig fig so v_i = 10 m/s

For problem 2), the elements "in play" are original velocity (aka initial velocity or v_i), final velocity (v_f), time (t), and distance (x). Equation #1 has those elements in it. Tweaking it by solving for time gives us t = 2(x_f - x_i)/(v_f + v_i). Again we can treat x_f - x_i as the total distance traveled. So t = 2(-2)/(-5 + 2) = -4/-3 = 1.33 sec. Again, there is only one sig fig in the givens so t = 1 sec

For problem 3), the elements "in play" are acceleration (a), original velocity (aka initial velocity or v_i), distance (x), and time (t). Equation #3 has those elements. We are asked to find distance and the equation is already set up for that so we use x_f = x_i + v_it + ½ at²

Notice we can consider xi = 0 so x_f is the distance we traveled from the start. So x_f = 0 + (3)(6) +½(3)(6²) = 0 + 18 + 54 = 72 meters. Again with only 1 sig fig, the distance traveled is 70 meters.