Question

jesse and mark are jogging along the route shown at a rate of 12 miles per hour. they start by jogging south along capitol street for 1 mile, then turn east on h street and jog for 1.75 miles. at that point, jesse is tired and decides to walk home along florida avenue at a rate of 5 miles per hour. mark plans to jog back the way they came. jesse wants to find out who will arrive home first and by how much time. which statements should he consider when solving the problem? choose five correct answers. mark will jog home a distance of 2.25 miles. jesse will walk home a distance of approximately 2.25 miles as found by (1^{2}+1.75^{2}=d^{2}) and then (2 + 3.0625=d^{2}). it will take mark about 0.23 hours to jog home as found by (2.75 = 12(t)). jesse will walk home a distance of approximately 2.0 miles as found by evaluating (1^{2}+1.75^{2}=d^{2}) and then (1+3.0625=d^{2}). mark will get home about 0.17 hours (or 10 minutes) sooner. mark will get home about 0.21 hours (or 12.5 minutes) sooner.

Explanation:

Step1: Calculate Mark's jogging distance

Mark jogs back the way they came. The total distance he jogs is \(1 + 1.75=2.75\) miles. So the statement "Mark will jog home a distance of 2.25 miles" is incorrect.

Step2: Calculate Mark's jogging time

Using the formula \(d = vt\) (where \(d\) is distance, \(v\) is velocity, and \(t\) is time), for Mark, \(d = 2.75\) miles and \(v = 12\) miles per hour. Then \(t=\frac{d}{v}=\frac{2.75}{12}\approx0.23\) hours.

Step3: Calculate Jesse's walking distance

Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 1\) mile and \(b = 1.75\) miles. Then \(d=\sqrt{1^{2}+1.75^{2}}=\sqrt{1 + 3.0625}=\sqrt{4.0625}\approx2.016\approx2.0\) miles. The calculation \(1^{2}+1.75^{2}=d^{2}\), then \(1 + 3.0625=d^{2}\) is correct for finding Jesse's distance.

Step4: Calculate Jesse's walking time

Using \(t=\frac{d}{v}\), for Jesse, \(d\approx2.0\) miles and \(v = 5\) miles per hour. Then \(t=\frac{2.0}{5}=0.4\) hours.

Step5: Calculate the time difference

The time difference \(\Delta t=0.4 - 0.23=0.17\) hours.

Answer:

• Jesse will walk home a distance of approximately 2.0 miles as found by evaluating \(1^{2}+1.75^{2}=d^{2}\) and then \(1 + 3.0625=d^{2}\).
• It will take Mark about 0.23 hours to jog home as found by \(2.75 = 12(t)\).
• Mark will get home about 0.17 hours (or 10 minutes) sooner.