Question

1. a father and his daughter go to the bus stop that is located 75 m from their front door. the father walks in a straight line while his daughter runs along a varied path. despite the different paths, they both end up at the bus stop at the same time. the fathers average speed is 2.2 m/s, and his daughters average speed is 3.5 m/s. (a) how long does it take the father and daughter to reach the bus stop? (b) what was the daughters total distance traveled? (c) if the daughter maintained her same average speed and traveled in a straight line like her father, how far beyond the bus stop would she have traveled?

a. (a) 21.43 s (b) 75 m (c) 0 m
b. (a) 21.43 s (b) 119 m (c) 44 m
c. (a) 34 s (b) 75 m (c) 0 m
d. (a) 34 s (b) 119 m (c) 44 m

Explanation:

Step1: Calculate time taken to reach bus - stop

We know that speed $v=\frac{d}{t}$, so $t = \frac{d}{v}$. The father's displacement $d = 75$ m and his speed $v_f=2.2$ m/s. The time taken $t$ is the same for both father and daughter. Using the father's data, $t=\frac{75}{2.2}\approx34$ s.

Step2: Calculate daughter's total distance

Since $v_d = 3.5$ m/s and $t = 34$ s, and $d=v\times t$, the daughter's total distance $d_d=3.5\times34 = 119$ m.

Step3: Calculate extra - distance if daughter walked straight

If the daughter walked in a straight - line like her father with speed $v_d = 3.5$ m/s for the same time $t = 34$ s, the distance she would cover is $d = 3.5\times34=119$ m. The straight - line distance to the bus stop is 75 m. So the extra distance beyond the bus stop is $119 - 75=44$ m.

Answer:

d. (a) 34 s (b) 119 m (c) 44 m