Question

The Bicycle race: Calculus Question...HELP! (see attatched)

Answer 1

I could do Q1 and Q2 but I am stuck on Q3...

Answer 2

Can anyone interpret q3? I'm not really sure what it is asking...

Answer 3

when the car and the biker meet, their velocities are equal, so would this expression be
\[k=v(car) \]
\[k=-t ^{2}+\frac{ 20 }{ 3 }t\]
is this right?

Answer 4

What did you do for the 1st part?

Answer 5

For the first part, I found the derivative of the distance function of the car (this is the velocity function of the car) \[d'=-t ^{2}+\frac{ 20 }{ 3 }t\]
then I found the acceleration function of the car (second derivative of the distance function
\[d''=-2t+\frac{ 20 }{ 3 }\]

But when the car reaches the biker, it moves at constant velocity, in other words, the car stops accelerating => \[d''=0\]
\[-2t+\frac{ 20 }{ 3 }=0\]
\[t=10/3\]

after this time, the car reaches amy and their velocities are equal => sub this into velocity function and I got k=100/9 m/s

Answer 6

Impressive!
Now in the 3rd part, if Carol was riding with some other velocity 'k' m/s, find the time when the car meets the bike. Given that the car starts from rest.

Answer 7

Okay, so does "car meets the bike" mean that their velocities are equal. therefore, \[k=v(car) \]
\[k=d'=-t ^{2}+20/3\]

Answer 8

When the car reaches the bike, they both will have travelled the same distance from the refreshment station.