# Problem Restatement A ball accelerates down a ramp at \( a = 1.30 \ \mathrm{m/s}^2 \) for \( t = 2.5 \) s. Find the length of the ramp (\( d \)), assuming it starts from rest. # Step-by-Step Solution ## 1. Identify the Kinematic Equation For constant acceleration, starting from rest (\( v_ = \)), displacement is: \[ d = v_ t + \frac{1}{2} a t^2 \] Since \( v_ = \): \[ d = \frac{1}{2} a t^2 \] ## 2. Plug in Values - \( a = 1.30 \) m/s² - \( t = 2.5 \) s \[ d = \frac{1}{2} \times 1.30 \times (2.5)^2 \] ## 3. Calculate Each Step - \( (2.5)^2 = 6.25 \) - \( \frac{1}{2} \times 1.30 = .65 \) - \( .65 \times 6.25 = 4.0625 \) \[ d = 4.0625 \ \mathrm{m} \] ## 4. Final Answer (with significant figures) Given the data (2 significant figures): \[ \boxed{4.1 \ \mathrm{m}} \] # Summary The length of the ramp is **4.1 m**.