Intro to Physics question?
Lee and Tom are two runners who often race each other. During their last race, Lee accelerated at 0.15m/s^2 from a standing start, while Tom accelerated at 0.40m/s^2, also from a standing start. After running for 30.0s, Tom fell down.
a)How far did Tom get before he fell down?
b)How far had Lee run when Tom fell down?
c)How fast was Lee running when Tom fell down?
d)How long after Tom fell did it take for Lee to catch up with Tom?
Update:i need to know how its done. I have the answers already
a)180m b)67.5m c)4.5m/s d)19s
More Questions From This User See All
Answers & Comments
Verified answer
You really only need one equation to solve this problem: the equation for uniform motion
X(t) =(a*t^2)/2+v0*t+x0
where t is time, a is acceleration, v0 is intial velocity, and x0 is intial position
Knowing this:
1.) For Tom, we know his a=.4, v0=0 (starting from rest) and x0=0 (the starting position)
so x(30)=(.4*30^2)/2 + 0*30+ 0 = 180m
2.) Same procedure for lee: a=.15, v0=0, x0=0
x(30)=(.15*30^2)/2 = 67.5m
3.) for this you must take the time derative of the above eqn, as the time deravitive of a position function is velocity:
v(t)= at+ v0
v0=0 and a = .15 for lee
v(30)= .15*30 + 0 = 4.5 m/s
4.) this one will take some algebra: We need the first eqn. again, accept this time, x0= 67.5m (the initial distance that tom is at after tom fell down) and v0=4.5m/s (the intial speed that lee is traveling at after tom fell down), and X(t)=180m (the distance that lee needs to travel to catch up to tom.) Solve for t:
Using the same eqn:
180= .15/2*t^2+4.5*t+67.5
Using the quadratic eqn to solve for t, you get t=18.99 or t=-78.99. Since the time can't be negative, the answer must be 18.99s or 19s.
Hope this helps
X = v initial x t + 1/2 a t^2 so V initial = 0 for both and X tom = 1/2 (.40) 30^2 = 180 m and X Lee = 1/2 (.15) 30^2 = 67.5 m
v final = vinitial + at so vf Lee = 0 + 0.15 x 30 = 4.5 m/s
Tom ran 180 m. For Lee this would mean 180 = 1/2 (.15) t^2 or t^2 = 2440 and t = 48.99 s so so it would take Lee 48.99 - 30 = 18.99 s to catch up to Tom
A really awesome equation to remember is velocity^2=initial velocity^2 + 2*acceleration*distance, and v=vo+at.
a) Plug in v=vo+at = 0 + 0.40*30= 12 m/s
Then use the first equation: 12^2 = 0^2 + 2* 0.40 * d and solve for d, to get 180 m.
b) Do the exact same thing, except with a=0.15.
c) You'd find this in step one of b).
d) You have the final distance (180 m) and the distance Lee had gotten so far (67.5 m). Find the difference between the two, 112.5 m, and that's how far Lee has to go. vo in this case= 4.5 m/s. acceleration is still 0.15 m/s^2. v^2=vo^2+2ad to find the final velocity he was at. Then plug that velocity, the initial velocity, and the acceleration into v=vo+at and solve for t.