A baseball player dives into third base with an initial speed of 7.9 m/s. If the coefficient of kinetic friction between the player and the ground is .41, how far does the player slide before coming to rest?
2007-08-06 03:13:37 · 4 answers · asked by Waverly Pascale 3 in Science & Mathematics ➔ Physics
He starts at 7.9 m/s and the force of friction is given by:
F = 9.8 m/s^2 * m * 0.41 = 4.02 * m N
His acceleration (deceleration) is F/m = 4.02 m/s^2
(His mass factors out - That's why it's not given in the problem.)
Thus, the time it takes for the player to go from 7.9 to 0 is:
t = Vo/a = 7.9 / 4.02 = 1.965 s
His speed graphed over time will be a triangle, and his distance covered will be the area underneath the triangle, so the distance covered will be given by:
d = 1/2 Vo * t = 1/2 * 7.9 * 1.965 = 7.76 m
2007-08-06 03:24:40 · answer #1 · answered by Garrett J 3 · 0⤊ 0⤋
Deceleration is the COF * the accel of gravity.
9.81m/s * .41 = 4.02m/s/s
distance = 1/2 deceleration X time^2
time is the amount of time it takes to go from 7.9 m/s to zero.
7.9 m/s -4.02m/s/s * time=0
7.9 = 4.02 *t
7.9/4.02 =t
t=1.97 seconds
distance = 1/2 (4.02) (1.97)^2 = 7.8 m/s
2007-08-06 03:35:08 · answer #2 · answered by eric l 6 · 0⤊ 0⤋
S = (7.9)^2/2*.41*9.8 = 7.77 m
2007-08-06 03:19:03 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋
hay, Gurl do your own homework. lol sike
idk i forgot took physics last year forgot now that what summer does to you
2007-08-06 03:28:05 · answer #4 · answered by Anonymous · 0⤊ 2⤋