25 October, 04:10

# Mark and Zoe are hiking a trail. Mark starts before Zoe. The expression 2t + 100 represents how far, in meters, Mark has hiked t seconds after Zoe starts, and 2t represents how far Zoe has hiked t seconds after starting. Is there a time during the hike when Mark and Zoe have hiked the same distance? Explain.

+1
1. 25 October, 04:19
0

Step-by-step explanation:

The distance hiked of Mark is represented by 2*t + 100

and the distance hiked of Zoe is represented by 2*t

here, you can see that the slope of both equations is equal, this means that in the same lapse of time, Zoe and Mark displace the same amount, but because Mark started earlier, he has a y-intercept bigger than zero, so for every value of t, the distance that Mark hiked will be higher than the one of Zoe.

We can represent this by:

2*t + 100 > 2*t

100 > 0

So the distance hiked by Mark is always bigger than the one of Zoe
2. 25 October, 04:20
0
Step-by-step explanation:

Given:

Mark distance in t seconds = 2t + 100

Zoe distance in t seconds = 2t

Velocity = distance/time

Therefore from the above equation,

Mark's velocity = 2 m/s

Zoe's velocity = 2 m/s

Distance covered by Mark = distance covered by Zoe

2t + 100 = 2t

If they both move at the same velocity but Mark starts 100 m before zoe. Therefore, there is no the same time that both will hike the same distance.

Mark will always be 100 m ahead of Zoe.