A bicyclist is finishing his repair of a flat tire when a friendrides by with a

Question

A bicyclist is finishing his repair of a flat tire when a friendrides by with a constant speed of 3.6m/s. Two seconds later the bicyclist hops on his bike andaccelerates at 2.2 m/s2until he catches his friend. (a) How much time does it take until he catcheshis friend?
1
Your answer differs from the correct answerby 10% to 100%. s
(b) How far has he traveled in this time?
2
Your answer differs from the correct answerby 10% to 100%. m
(c) What is his speed when he catches up?
3
Your answer differs from the correct answerby 10% to 100%. m/s
(a) How much time does it take until he catcheshis friend?
1
Your answer differs from the correct answerby 10% to 100%. s
(b) How far has he traveled in this time?
2
Your answer differs from the correct answerby 10% to 100%. m
(c) What is his speed when he catches up?
3
Your answer differs from the correct answerby 10% to 100%. m/s

Explanation / Answer

So we can call the friend who rides by.. cyclist 1 , heis moving at constant speed. We can call the one who jumps on two seconds later, Cyclisttwo... For cyclist 1, we have X = Xo + Vx*t Since two seconds passes by until the other cyclist comes intoquestion, lets call Xo (2sec*3.6m/s) = 7.2m
X =7.2 + 3.6 *t
For Cyclist 2, we have X = Xo + VoT + 1/2at^2 X = 0 + 0*t + 1/2(2.2)*t^2 set these two expressions equal, so their x will match up ,thats when theyve caught up to one another.. 7.2 + 3.6t = 1.1t^2 1.1t^2 - 3.6t - 7.2 = 0 solve quadratic.. 4.67 sec, t = -1.4 sec, but time cant be negative so t = 4.67secs.. Plug this time in either equation and well get the totaldistance...and they should match up ( theyve caught up,remember?) X = 7.2 + 3.6(4.67) = 24.012 m X = 1.1(4.67)^2 = 24m Speed when he catches up is V = Vo + At V = (2.2)(4.67) V = 10.27 m/s when he catches up

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