Problem 4-12 - Vector components - Part 6 - (B)

A car is travelling along a winding highway at a constant speed of \(82\; km/h.\) At 3:00 p.m., the car is heading \(38.2^\circ\) east of north, and at 3:15 p.m. it is heading \(12.7^\circ\) south of east. Define the \(+x\) and \(+y\) directions to be east and north, respectively. Use Eq. 4-8 to determine the \(x\)- and \(y\)-components of the average acceleration during this time interval.  

Accumulated Solution

\(82\; km/hr = 22.8\; m/s\)

\(v_{fx} = 22.8 \cos 12.7^\circ = 22.24\; m/s \\ v_{fy} = -22.8 \sin 12.7^\circ = -5.01\; m/s \\ v_{ix} = 22.8 \sin 38.2^\circ = 14.10\; m/s \\ v_{iy} = 22.8 \cos 38.2^\circ = 17.92\; m/s\)

Correct!

The acceleration in the \(x - \)direction is:

(A)   \(a_x = (22.24 - 14.10)/900 = 9.0\times 10^{-3} m/s^2\)

(B)   \(a_x = (14.10 - 22.24)/900 = -9.0 \times 10^{-3} m/s^2\)

(C)   \(a_x = (22.24 - 5.01)/900 = 0.019\; m/s^2\)