A car starts from rest and accelerates uniformly over a time of 11 seconds for a distance of \( 144 \mathrm{~m} \). Determine the acceleration of the car. \( \left(m / s^{2}\right. \) unit
Answer
3. Substitute the known values and solve for acceleration: \(144 = 0(11) + \frac{1}{2}a(11)^2 \Rightarrow a = \frac{288}{(11)^2} \Rightarrow a ≈ 2.3766~\mathrm{m/s^2}\)
Step 1 ：1. Identify the knowns and unknowns: initial velocity \(v_0 = 0\mathrm{~m/s}\), time \(t = 11\mathrm{~s}\), distance \(d = 144\mathrm{~m}\), and we need to find acceleration \(a\).
Step 2 ：2. Choose the appropriate kinematic equation: \(d = v_0t + \frac{1}{2}at^2\).
Step 3 ：3. Substitute the known values and solve for acceleration: \(144 = 0(11) + \frac{1}{2}a(11)^2 \Rightarrow a = \frac{288}{(11)^2} \Rightarrow a ≈ 2.3766~\mathrm{m/s^2}\)
