At \(t=0,\) an automobile traveling north begins to make a turn. It follows one- quarter of the arc of a circle with a radius of \(10.0 \mathrm{m}\) until, at \(t=1.60 \mathrm{s},\) it is traveling east. The car does not alter its speed during the turn. Find (a) the car's speed, (b) the change in its velocity during the turn, and (c) its average acceleration during the turn.

To find the distance traveled along the arc, we'll use the formula for the length of a circular arc: \(s = r \theta,\) where \(s\) is the arc length, \(r\) is the radius, and \(\theta\) is the angle in radians. Since the car follows one-quarter of the arc, it covers an angle of \(\frac{\pi}{2}\) radians. Therefore, \(s = 10\cdot \frac{\pi}{2}= 5\pi \mathrm{m}\)

Now that we know the distance traveled, we can find the car's speed by dividing the distance by the time it took to complete the turn: \(v = \frac{s}{t} = \frac{5\pi \mathrm{m}}{1.60 \mathrm{s}} = \frac{5}{1.60}\pi \mathrm{\frac{m}{s}}\)

The initial velocity is northward, and the final velocity is eastward. We know the magnitude of the velocity (from the previous step), so we can represent the initial and final velocities as vectors: Initial velocity: \(\vec{v_i} = (0, \frac{5}{1.60}\pi \mathrm{\frac{m}{s}})\) Final velocity: \(\vec{v_f} = (\frac{5}{1.60}\pi \mathrm{\frac{m}{s}}, 0)\) Now, we will find the change in velocity by subtracting the initial velocity from the final velocity: \(\Delta \vec{v} = \vec{v_f} - \vec{v_i} = (\frac{5}{1.60}\pi \mathrm{\frac{m}{s}}, -\frac{5}{1.60}\pi \mathrm{\frac{m}{s}})\)

To find the average acceleration, we need to divide the change in velocity by the time it took to complete the turn: \(\vec{a}_{avg} = \frac{\Delta \vec{v}}{t} = \frac{(\frac{5}{1.60}\pi \mathrm{\frac{m}{s}}, -\frac{5}{1.60}\pi \mathrm{\frac{m}{s}})}{1.60 \mathrm{s}} = (\frac{5\pi}{(1.60)^2} \mathrm{\frac{m}{s^2}}, -\frac{5\pi}{(1.60)^2} \mathrm{\frac{m}{s^2}})\) So, we have found: (a) the car's speed: \(\frac{5}{1.60}\pi \mathrm{\frac{m}{s}}\), (b) the change in its velocity during the turn: \((\frac{5}{1.60}\pi \mathrm{\frac{m}{s}}, -\frac{5}{1.60}\pi \mathrm{\frac{m}{s}})\), (c) its average acceleration during the turn: \((\frac{5\pi}{(1.60)^2} \mathrm{\frac{m}{s^2}}, -\frac{5\pi}{(1.60)^2} \mathrm{\frac{m}{s^2}})\).