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 initial velocitytoward 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, . Let the time at which the dragster starts to accelerate be t=0

a)Find numerical values for t max and D start in seconds and meters for the (reasonable) values initial velocity (26.8 m/s) and a=50m/s2

Explanation / Answer

No collision can occur if the dragster has greater speed than the speed of the car behind it. 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*t^2 in order for sd(t)>sc(t) v0*t-d>.5*a*t^2 if you allow the car to just touch the dragster 0=d-v0*t+.5*a*t^2 or 0=2*d-2*v0*t-a*t^2 using the quadratic equation t=(2*v0+/-(sqrt(4*v0^2-4*a*2*d))/(2*a) If the dragster does not get touched by the car 2*a*d>v0^2 if it just touches the dragster 2*a*d=v0^2 =:> d = v0^2/2a and t=v0/a Vd (tmax) = Vo t max = Vo/ a t max = 26.8/50 t max = 0.536 sec = 0.54 secs ---> Answer D car = v0^2/2a = (26.8)^2 /2*50 = 7.182 m = 7.2 m ---> Answer

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