I know its asking for a lot
I really need help with this physics question. The answer to the question is 2.89m/s/s, but I don't know how to arrive at that answer. I've tried like a ton of times and my answer is never what the sheet says it is. I'm beginning to think my teacher made an error (I got the answer 11.84m/s/s)
Two sprinters are in a 100m race. When the starting gun sounds, "A" accelerates at 4.5m/s/s for 3 seconds and then maintains a constant velocity for the rest of the race. "B" stumbles at the line and doesn't begin to accelerate until .75 seconds after the gun. Assuming "B" accelerates the whole way, how fast does she have to accelerate to catch "A" at the 80m mark?
Here are some equations that might be helpful:
delta x= v initial+0.5at^2
v=delta x/delta t
a= delta v/ delta t
v final= v initial+at
vfinal^2=vinitial^2+(2*a*deltax)
2 Comments • Newest first
Haha, I don't understand any of this.....
Both of you are wrong.
Firstly lets imagine it's a 80m race. Let us work out d, the distance sprinter A travels in uniform acceleration.
d = 1/2 a t^2
d= 1/2 x 4.5 x 3^2
d = 20.25m
Now lets find how long it takes for him to finish the race. i.e reach 80m.
Distance he travels at constant velocity = 80 -20.25= 59.75m
He travels at the speed 4.5 x 3 = 13.5m/s
Time taken = 59.75/13.5 = 4.4259s
So total time taken = 4.4259 + 3 = 7.4259s
So therefore, sprinter B must do 80m in 7.4259 - 0.75 = 6.6759 s due to the 0.75 seconds he lost stumbling.
Using formula for constant acceleration:
d= 1/2 a t^2
80 = 1/2 a 6.6759^2
160/ 6.6759^2 = a
a= 3.59ms^-2
EDIT: User above me is wrong, please ignore his advice. Both you, your teacher, and the person who posted above me are all wrong.