A sailboat starts from rest and accelerates at a rate of 0.23 m/s2 over a distance of 406 m.
(a) Find the magnitude of the boat's final velocity.
(b) Find the time it takes the boat to travel this distance.
2007-09-09 06:30:16 · 3 answers · asked by ? 1 in Science & Mathematics ➔ Physics
b) x = x_0 + v_0 * t + 1/2*a*t²
406 = 1/2*.023t²
t = 59.42 s
a) v(t) = ∫ 0.23 dt
v(t) = 0.23t
v(59.42s) = 13.67m/s
2007-09-09 06:41:10 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The velocity is v = at and distance is d = 1/2 a t^2 since we're starting from rest.
So first we get t by using the second eq since we know everything in it except t: 406 = 1/2 * .23 t^2 or
t^2 = 789 or t = 28.1 sec.
Now use the first eq.
v = .23 * 28.1 = 6.5 m/s Which is pretty fast. Check my numbers!
2007-09-09 13:41:36 · answer #2 · answered by rrsvvc 4 · 0⤊ 0⤋
a. v = at
s = (1/2)at^2 or t = SQRT(2s/a)
v = aSQRT(2s/a)
v = SQRT(2sa)
a = 0.23 m/s^2, s = 406 m
v = SQRT(2*406*0.23) = SQRT(186.76)
v = 13.67 m/s
b. v = at or t = v/a
t = 13.67/0.23
t = 59.42 seconds
2007-09-09 13:45:31 · answer #3 · answered by Captain Mephisto 7 · 0⤊ 0⤋