Long Rubber Hammer
Long Rubber Hammer
Check out this page if you are looking for Long Rubber Hammer
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 027M Long Blk Nonslip Handle Rubber Mallet Hammer Tool  US $12.54
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 TOOLS TOOL RUBBER MALLET 12 LONG HAMMER US $9.99
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 Home Garden Black 025M Long Rubber Mallet Hammer Tool US $17.28
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 Home Hand Tool 118 Long Rubber Mallet Hammer Black Clear Yellow US $21.12
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 Metal Silver Hammer w Rubber Grip Handle 12 1 2 Long US $14.99
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 Soft Tap 10 Long Antiskid Double Head Rubber Mallet Hammer US $9.22
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 Hand Tool 11 Long Rubber Mallet Hammer Black Clear Yellow US $14.11
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 031M Long Antiskid Handle Black Rubber Mallet Hammer US $25.29
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Physics Problem I don't know how to start with this?
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.
What is the longest time (tmax)after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?
I don't understand how to find a common time between the two given two different velositys. Can anyone help?
Let's say at t = 0 you are 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 between you is then is s = d - v0*t + 0.5*a*t². You will collide s = 0,
0.5*a*t² - v0*t + d = 0
t = v0 ± √[v0² - 2*a*d] / a
t = v0/a ± √[(v0/a)² - 2*d/a]