Question
Car $A$ is traveling on a highway at a constant speed $\left(v_{\lambda}\right)_{0}=60 \mathrm{mi} / \mathrm{h}$ and is $380 \mathrm{ft}$ the entrance of an access ramp when car $B$ enters the acceleration lane at that point at a speed $\left(v_{B}\right)_{0}=15 \mathrm{mi} / \mathrm{h}$. Car $B$ accelerates uniformly and enters the main traffic lane after traveling $200 \mathrm{ft}$ in $5 \mathrm{s}$. It then continues to accelerate at the same rate until it reaches a speed of $60 \mathrm{mi} / \mathrm{h},$ which it then maintains. Determine the final distance between the two cars.
Step 1
We know that there are 5280 feet in one mile and 3600 seconds in one hour. So, the speed of car B when it enters the acceleration lane is $v_{B0} = 15 \times \frac{5280}{3600} = 22 \, \text{ft/s}$. Show more…
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