(1) Straight to the Point Direction: Solve the following and SHOW your complete solution. (3 points each) (Note: Use ONLY the back side of this paper for your solution and the graph A rocket is launched with an initial velocity of \( 50 \mathrm{~m} / \mathrm{s} \) and accelerates uniformly at \[ a(t)=20 \mathrm{~m} / \mathrm{s}^{2} \text { for } 4 \text { seconds. } \] (1-3) Approximate the total change in velocity over the 4 -second interval using a Riemann sum with \( n=5 \) subintervals, considering left endpoints, right endpoints, and midpoints. Let: \( 0 \leq t \leq 4 \) and \( n=5 \). (4-5) Write the definite integral representing the total change in velocity and evaluate it.

To approximate the total change in velocity using a Riemann sum with \( n=5 \) subintervals over the interval \( 0 \leq t \leq 4 \) seconds, we will first determine the width of each subinterval, \( \Delta t \), which is given by: \[ \Delta t = \frac{(4 - 0)}{5} = \frac{4}{5} = 0.8 \text{ seconds.} \] Now, for the left endpoints, we will calculate the change in velocity at the following time points: \( t = 0, 0.8, 1.6, 2.4, 3.2 \). The acceleration function is constant, \( a(t) = 20 \, \text{m/s}^2 \). Therefore, for each subinterval, the change in velocity \( \Delta v \) can be calculated as: \[ \Delta v = a(t) \cdot \Delta t = 20 \, \text{m/s}^2 \cdot 0.8 \, \text{s} = 16 \, \text{m/s}. \] So, for the left endpoint Riemann sum: \[ \text{Total change in velocity} = 5 \cdot 16 \, \text{m/s} = 80 \, \text{m/s}. \] For the right endpoints, we will evaluate the change in velocity at the points: \( t = 0.8, 1.6, 2.4, 3.2, 4.0 \). The calculation remains the same (since \( a(t) \) is constant): The right endpoint Riemann sum will also yield: \[ \text{Total change in velocity} = 5 \cdot 16 \, \text{m/s} = 80 \, \text{m/s}. \] Now for the midpoint Riemann sum, we need to evaluate at the midpoints \( t = 0.4, 1.2, 2.0, 2.8, 3.6 \). Again, the calculation will yield: \[ \Delta v = 20 \, \text{m/s}^2 \cdot 0.8 \, \text{s} = 16 \, \text{m/s}, \] and thus: \[ \text{Total change in velocity} = 5 \cdot 16 \, \text{m/s} = 80 \, \text{m/s}. \] Now, for the definite integral representing the total change in velocity, we can express it as: \[ \int_0^4 a(t) \, dt = \int_0^4 20 \, dt. \] Evaluating the integral: \[ \int_0^4 20 \, dt = 20 \cdot [t]_0^4 = 20 \cdot (4 - 0) = 80 \, \text{m/s}. \] So, the total change in velocity over the 4-second interval is \( 80 \, \text{m/s} \).