Question

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.0 m/s. Two seconds later the bicyclist hops on his bike and accelerates at 2.0 m/s2 until he catches his friend.
a.) How much time does it take until he catches his friend?
b.) How far has he traveled in this time?
c.) What is his speed when he catches up?

Answer 1

so t+3/2 but add 2 for total time making it 7/2s. then multiply by 3 to get 10.5 for distance. yes?

Answer 2

t=3/2. typo

Answer 3

but how do I get speed?

Answer 4

Wait !

Answer 5

Aight, so let's call the guy repairing his tire b, and his friend will be f. Now, at t=0, his friend rides by at 3.6m/s. Therefore, the position of his friend x\(_f\) = 0 + 3.6t, assuming that the position where b was fixing his bike x = 0. Now, the other guy starts accelerating at a\(_b\) = 2.8 m/s^2. Now, from the standard kinematic equation with acceleration:

x\(_b\) = x\(_0\) + v0b*t + 0.5*ab*t^2

However, x0=0, v0b=0, and we'll replace t with tb = t-2

xb = 0.5*ab*(t-2)^2

We want to know when xb = xf:

3.6t = 0.5*(2.8)*(t-2)^2
3.6t = 1.4*(t-2)^2
3.6t = 1.4*(t^2-4t+4)
3.6t = 1.4t^2-5.6t+5.6
0 = 1.4t^2-(5.6+3.6)t+5.6
0 = 1.4t^2-9.2t+5.6

Now we apply the quadratic equation:

t = [9.2 +/- sqrt(9.2^2-4(1.4)(5.6)]/2(1.4)
t1 = 5.893
t2 = 0.679

The guy who was fixing his bike didn't start riding until t=2, so it has to be t1 = 5.893. We'll double-check:

xf = 0 + 3.6t = 3.6(5.893) = 21.215
xb = 0.5*ab*(t-2)^2 = 1.4(t-2)^2 = 1.4(3.893)^2 = 21.218

Let's call that close enough! So we know it took him 5.893 seconds from when his friend walked by or 3.893 seconds after he started riding, and he traveled 21.21 meters in that time.

For the final part of the problem:

vb = v0 + ab*tb
vb = 0 + 2.8*(t-2)
vb = 2.8(3.893)
vb = 10.90 m/s

He passes his friend going 10.9 m/s (which is pretty insanely fast btw

Answer 6

Getting it ?

Answer 7

hold on. its going to take a minute to read it all

Answer 8

yes. that helps so much. thank you

Answer 9

\(\huge\sf Do~you~know? \\ \huge Now~you~can~easily~search~users~\\ \huge and~questions~on~openstudy!\\ \huge Please \ visit ~\href{//openstudyab.comoj.com/search.html}{this} \huge \ utility~website~to~try~it\\ \huge now. Don't~forget~to~provide~us ~feedback.\)

Answer 10

\(\color{red}{\huge\bigstar}\huge\text{You're Most Welcome! }\color{red}\bigstar\)
\(~~~~~~~~~~~~~~~~~~~~~~~~~~~\color{green}{\huge\ddot\smile}\color{blue}{\huge\ddot\smile}\color{pink}{\huge\ddot\smile}\color{red}{\huge\ddot\smile}\color{yellow}{\huge\ddot\smile}\)