A sailboat starts from rest and accelerates at a rate of 0.18 m/s2 over a distance of 494 m.
(a) Find the magnitude of the boat's final velocity.
(b) Find the time it takes the boat to travel this distance.
any physics teacher can help me ??????
A car starts from rest and travels for 4.1 s with a uniform acceleration of +1.4 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.8 m/s2. The breaks are applied for 1.80 s.
Vf^2=2*acceleration*distance
distance=1/2 a*t^2 solve for time t.
1. V^2 = Vo^2 + 2a*d.
a. V^2 = 0 + 0.36*494, V = ?.
b. d = 0.5a*t^2.
494 = 0.5*0.18*t^2. t = ?.
2. V = Vo + a*t = 0 + 1.4*4.1 = 5.74 m/s.
V^2 = Vo^2 + 2a*d.
0 = (5.74)^2 - 3.6d. d = ?.
87.01
To find the answers to these questions, we can use the equations of motion. The equations we need to use are:
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as
where:
v is the final velocity
u is the initial velocity (in this case, it is 0 m/s since the sailboat starts from rest)
a is the acceleration (0.18 m/s^2)
t is the time taken
s is the distance traveled (494 m)
(a) To find the magnitude of the boat's final velocity, we need to use equation 1. Since the initial velocity u is 0, the equation simplifies to:
v = at
Substituting the values, we have:
v = 0.18 m/s^2 * t
(b) To find the time it takes for the boat to travel the given distance, we need to use equation 2, which is:
s = ut + (1/2)at^2
Rearranging the equation, we get:
(1/2)at^2 + ut - s = 0
This is a quadratic equation in terms of t. We can solve this equation using the quadratic formula:
t = (-u ± √(u^2 + 2as)) / a
Substituting the values, we have:
t = (-0 ± √(0^2 + 2 * 0.18 * 494)) / 0.18
Simplifying the equation will give us the answer for part (b).
Now, let's solve these equations to find the final velocity and time taken.