What is its displacement (in meters) from t1 to t2? Keep 3 digits after the decimal point.

Question

At t1 = 5.0s an object's velocity is v1 = 8.790m/s, and at t2 = 11.0s its velocity is v2 = 17.730m/s. Assuming a constant acceleration.

Answer 1

The shortest way to solve will be use kinematic equation: v₂² - v₁²=2a(∆x).

Hence, at first you need to find acceleration, as it is uniform, a=∆v/∆t=(17.73-8.79)/(11.0 -5.0)=1.49 m/s²;

 Then, using kinematic equation   v₂² - v₁²=2a(∆x), and solving it for ∆x=(v₂² - v₁²)/2a =79.56 m

Answer 2

Notice that since we have a constant acceleration, we can apply the following kinematics formula:

Δx = 1/2 * (v_i + v_f) * Δt

Plugging in our, values, we find

Δx = 1/2 * (8.790 + 17.730) * (11 - 5) = 79.560 m.

Intuitively, you can think of the formula as: displacement equals the average velocity multiplied by the amount of time traveled. To derive it, we can start with the general equation of motion:

Δx = v_iΔt + 1/2 * a(Δt)².

Since acceleration is simply the change in velocity over change in time, or a = Δv/Δt = (v_f - v_i)/Δt, we find

Δx = v_iΔt + 1/2 * (v_f - v_i)Δt

= (1 - 1/2) * v_iΔt + (1/2) * v_fΔt

= 1/2 * (v_i + v_f) * Δt

Then we can just rearrange the terms to get us to the equation we used for our problem. Importantly, this works because we know our acceleration is constant!