Crash and Collision - Kinematics Question?

Question

Vancouver is 300 km away from Seattle. Two friends, one leaving from each city on a bike want to find a campsite between the two cities. One cyclist starts from vancouver at 25 km/h. His friend will start 2.0h later from seattle at 32 km/h. How far from Vancouver do the friends meet?

Attempt:

dA + dB = 300

vA(t) + vB(t+2) = 300

25t + 32t + 2 = 300

t = 4.14 s

now how do i find the distance travelled by each cyclist?

not sure if this is even the correct direction, but any help would be greatly appreciated! Thank you!!

btw- correct answer is 160 km

Answer 1

To stay consistent with your labelling, write the position of the Vancouver cyclist as $ \ x_A = 25 \ (t+2) \ $ and the position of the Seattle cyclist as $ \ x_B = 300 - 32t \ . $ We are calling the location of Vancouver $ \ x = 0 \ $ by doing so. Remember that the Vancouver cyclist is the one who has been pedalling two hours longer.

You want to solve for the time when $ \ x_A = x_B \ . $ That time $ \ T \ $ can then be entered into either position function to find the distance from Vancouver where they meet.

Note: the answer rounds off to 160 km., in case you are wondering about the "ugly" fractions.

Answer 2

Hint: Suppose that the friends begin cycling at the same time, but instead of starting from $300$ kilometers away from each other, they start $250$ kilometers away from each other. If $x$ is the answer to this equation, the answer to your original equation will be $x + 50$.

Answer 3

Your direction is fine. It is good to define your variables, like $dA$ is the distance traveled by the one from Vancouver, etc. The vB should be multiplied by $(t-2)$ as the one from Seattle leaves two hours later, so that cyclist is traveling two hours less. You dropped multiplying the $2$ by vB, so it should become $64$ (with a minus sign). The unit on $t$ should be hr, not s. One cyclist traveled $vA(t)$ and the other $vB(t-2)$