Question

To demonstrate the tremendous acceleration of a top fuel dragracer, you attempt to run your car into the back of a dragster thatis "burning out" at the red light before the start of a race.(Burning out means spinning the tires at high speed to heat thetread and make the rubber sticky.)
You drive at a constant speed of v0 toward the stopped dragster, not slowingdown in the face of the imminent collision. The dragster driversees you coming but waits until the last instant to put down thehammer, accelerating from the starting line at constant
acceleration,a. Let the time at which thedragster starts to accelerate be t=0.
A) What is tmax, the longest time after thedragster begins to accelerate that you can possibly run into theback of the dragster if you continue at your initial velocity?
B) Assuming that the dragster has started at the last instantpossible (so your front bumper almost hits the rear of thedragster at t = tmax), find your distance from thedragster when he started. If you calculate positions on the way tothis solution,
choose coordinates so that the positionof the drag car is 0 at t = 0. Remember that you are solvingfor a distance (which is a magnitude, and can never be negative),not a position (which can be negative).

Answer 1

Answer:

a. 2v₀/a   b. 2v₀/a  

Explanation:

a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.

Since the dragster starts from rest with an acceleration, a, using

s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster

s' = 0t + 1/2at²

s' = 1/2at²

Since the distance moved by me and the dragster must be the same,

s = s'

v₀t. =  1/2at²

v₀t. - 1/2at² = 0

t(v₀ - 1/2at) = 0

t= 0 or v₀ - 1/2at = 0

t= 0 or v₀ = 1/2at

t= 0 or t = 2v₀/a  

So the maximum time tmax = 2v₀/a

b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s =  v₀(2v₀/a)

= 2v₀/a