The figure shows the velocity function for motion along a line. Assume the motion begins with an initial position of \( s(0)=0 \). Determine the following. a. The displacement between \( \mathrm{t}=0 \) and \( \mathrm{t}=10 \). b. The distance traveled between \( \mathrm{t}=0 \) and \( \mathrm{t}=10 \). c. The position at \( \mathrm{t}=10 \). a. The displacement is \( \frac{9}{2} \) unit(s). (Simplify your answer. Type an integer or a fraction.) b. The distance traveled between \( \mathrm{t}=0 \) and \( \mathrm{t}=10 \) is \( \frac{27}{2} \) unit(s). (Simplify your answer. Type an integer or a fraction.) c. The position at \( \mathrm{t}=10 \) is \( \mathrm{s}(10)=\square \). (Simplify your answer. Type an integer or a fraction.)

If we assume that the velocity function is given graphically and you're looking for position and distance traveled, you can find the answers with some careful calculations. To find the position at \( t = 10 \), use the formula for displacement, which was already found as \( \frac{9}{2} \). Since you started from an initial position of \( s(0) = 0 \), simply add the displacement to it: \[ s(10) = s(0) + \text{displacement} = 0 + \frac{9}{2} = \frac{9}{2} \] Now, you need to compute the distance traveled. Since the distance involves the total movement regardless of direction, you'll consider the absolute values of displacement over the interval. You already noted that the distance traveled is \( \frac{27}{2} \) units. Therefore, to wrap it all up: c. The position at \( t = 10 \) is \( s(10) = \frac{9}{2} \) units.