two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2 is moving to the left with speed V sub 0. Car 1 begins to move at t=0, speeding up with a constant acceleration a. Car 2 continues to move with constant velocity. A.) At what time do the two cars collide? B.) Find the speed of car 1 just before it collides with car 2.
2006-07-30 19:57:46 · 3 answers · asked by just_askin' 1 in Science & Mathematics ➔ Physics
how can i find the t? what is the formula... i am really weak when it comes to deriving a formula.. can someone teach me the tricks? sigh...
2006-07-30 20:12:56 · update #1
Say that the time taken is t seconds.
Car one travels
S=ut + 1/2 a t^2
= 0 + 1/2 a t^2 meters
Car Two travels
S = ut + 1/2 at^2
= Vt
These two together travel D meters
Therefore (adding these distances)
Vt + 1/2 a t^2 = D
a t^2 + 2Vt - D = 0
THis is a quadratic equation whose roots are t1 and t2 and can be obtained by applying the formula for quadratic's roots.
t1 = (-2a + root(4V^2 +4aD) / 2a
t2 = (-2a - root(4V^2 +4aD) / 2a
Of this t1 is positive and gives the time.
Velocity of Car 1
v = u + at
= 0 + a* (-2a + root(4V^2 +4aD) / 2a
= -a + root(V^2 + ad)
Hope this helps
2006-07-30 20:33:58 · answer #1 · answered by blind_chameleon 5 · 0⤊ 0⤋
D=V*t+(a*t^2)/2
You will get t from here..
Speed of car 1=a*t
2006-07-31 03:03:33 · answer #2 · answered by Synaps 2 · 0⤊ 0⤋
Do your own hw.
2006-07-31 03:37:08 · answer #3 · answered by Neil 2 · 0⤊ 0⤋