To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt

Question

To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster that is "burning out" at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.) You drive at a constant speed of v0 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, a. Let the time at which the dragster starts to accelerate be t=0.

Part A What is tmax, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?

Assuming that the dragster has started at the last instant possible (so your front bumper almost hits the rear of the dragster at t= tmax), find your distance from the dragster when he started. If you calculate positions on the way to this solution, choose coordinates so that the position of the drag car is 0 at t=0. Remember that you are solving for a distance (which is a magnitude, and can never be negative), not a position (which can be negative).

ind numerical values for tmax and Dstart in seconds and meters for the (reasonable) values v0=60mph (26.8 m/s) and a=50m/s2. Separate your two numerical answers by commas, and give your answer to two significant figures.

please showed all work including. formulas and steps

Explanation / Answer

Consider t = 0 and we are at a distance d behind the racer. From that time on, you advance at a rate v0*t, while the racer advances at a rate 0.5*a*t2.

The distance , s = d - v0*t + 0.5*a*t2. You will collide s = 0,

0.5*a*t2 - v0*t + d = 0

t = v0 ± sqrt[v2 - 2*a*d] / a

t = v0/a ± sqrt[(v/a)2 - 2*d/a]

The car is moving at a constant speed directly at the back of the dragster
At t=0, speed is v0 and distance is d
the equations of motion for the car are
sc(t)=v0*t-d

for the dragster, the equations of motion are
sd(t)=.5*a*t2

in order for sd(t)>sc(t)
v0*t-d>.5*a*t2

if you allow the car to just touch the dragster
0=d-v0*t+.5*a*t2
or
0=2*d-2*v0*t-a*t2
using the quadratic equation

t=(2*v0+/-(sqrt(4*v02-4*a*2*d))/(2*a)
If the dragster does not get touched by the car
2*a*d>v02
if it just touches the dragster
2*a*d=v02
and
t=v0/a

check
2*a*d/a-d=d

.5*2*a*d/a=d

checks

so
d=v02/(2*a)

if

v0=26.8 and a=50

d=26.82/(10)
71.82 m

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