Question

a woman walks at 2.1 m/s due east. after 199 seconds, she changes her direction by 20° to the south at 2.9 m/s for 187 seconds. what is the distance between where she started and where she ended?
946 m
835 m
577 m
790 m

Explanation:

Step1: Calculate the first - part distance

Use the formula \(d = vt\). For the first part, \(v_1=2.1\space m/s\) and \(t_1 = 199\space s\). So \(d_1=v_1t_1=2.1\times199 = 417.9\space m\) (in the east - direction).

Step2: Calculate the second - part distance components

For the second part, \(v_2 = 2.9\space m/s\) and \(t_2=187\space s\). So \(d_2=v_2t_2=2.9\times187 = 542.3\space m\).
The east - component of \(d_2\) is \(d_{2x}=d_2\cos(20^{\circ})\) and the south - component is \(d_{2y}=d_2\sin(20^{\circ})\).
\(d_{2x}=542.3\times\cos(20^{\circ})\approx542.3\times0.9397 = 509.2\space m\), \(d_{2y}=542.3\times\sin(20^{\circ})\approx542.3\times0.3420 = 185.5\space m\).

Step3: Calculate the total east - component and use the Pythagorean theorem

The total east - component \(D_x=d_1 + d_{2x}=417.9+509.2 = 927.1\space m\).
The south - component \(D_y = d_{2y}=185.5\space m\).
By the Pythagorean theorem \(D=\sqrt{D_x^{2}+D_y^{2}}\).
\(D=\sqrt{927.1^{2}+185.5^{2}}=\sqrt{859414.41 + 34410.25}=\sqrt{893824.66}\approx946\space m\)

Answer:

946 m