A ball rolling with initial velocity of 40m/s(W) undergoes an acceleration of 5.0m/s^2(N) for a period of 6.0seconds.
(A) what is the final velocity of the ball?
(B) what is the displacement of the ball in the 6.0 sec
(A) Well, the ball must have heard the phrase "go big or go home" because it definitely went big with that initial velocity! Now, let's figure out its final velocity. We can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time. Plugging in the values we have, we get v = 40 + (5 x 6) = 70 m/s. So, the final velocity of the ball is 70 m/s.
(B) Ah, the old displacement question, always making me ponder life's big mysteries. To find the displacement, we need to use the formula s = ut + 1/2at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is time. Plugging in the given values, we get s = (40 x 6) + (1/2 x 5 x (6^2)) = 240 + 90 = 330 m. So, the displacement of the ball in the 6 seconds is 330 meters.
To find the final velocity of the ball (A), you can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Plugging in the given values:
v = 40 m/s (west) + 5.0 m/s^2 (north) * 6.0 s
First, we need to convert the initial velocity into a vector form by expressing it as 40 m/s (west) = -40 m/s (east).
So, we have:
v = -40 m/s + 5.0 m/s^2 * 6.0 s
Calculating:
v = -40 m/s + 30 m/s = -10 m/s (east)
Therefore, the final velocity of the ball is 10 m/s (east).
To find the displacement of the ball in the 6.0 seconds (B), you can use the equation:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity
a = acceleration
t = time
Plugging in the given values:
s = -40 m/s * 6.0 s + (1/2) * 5.0 m/s^2 * (6.0 s)^2
Calculating:
s = -240 m + (1/2) * 5.0 m/s^2 * 36 s^2
s = -240 m + 90 m
s = -150 m
Therefore, the displacement of the ball in the 6.0 seconds is 150 m (west).
To find the answers to the given questions, we can use the equations of motion. There are four equations of motion:
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as
4. s = vt - (1/2)at^2
where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
- s is the displacement
(A) To find the final velocity (v):
Using equation 1, we have:
v = u + at
Given:
Initial velocity (u) = 40 m/s (West)
Acceleration (a) = 5.0 m/s^2 (North)
Time (t) = 6.0 seconds
Substituting the values into the equation:
v = 40 m/s (West) + 5.0 m/s^2 (North) * 6.0 seconds
The initial velocity and acceleration are in different directions, so we need to subtract them:
v = 40 m/s - 30 m/s
The final velocity is:
v = 10 m/s (West)
Therefore, the final velocity of the ball is 10 m/s to the west.
(B) To find the displacement (s):
Using equation 2, we have:
s = ut + (1/2)at^2
Given:
Initial velocity (u) = 40 m/s (West)
Acceleration (a) = 5.0 m/s^2 (North)
Time (t) = 6.0 seconds
Substituting the values into the equation:
s = 40 m/s (West) * 6.0 seconds + (1/2) * 5.0 m/s^2 (North) * (6.0 seconds)^2
First, calculate the displacement component due to initial velocity:
s1 = 40 m/s * 6.0 seconds
Next, calculate the displacement component due to acceleration:
s2 = (1/2) * 5.0 m/s^2 * (6.0 seconds)^2
Finally, subtract s2 from s1 to get the total displacement:
s = s1 - s2
Calculate the values:
s1 = 240 m
s2 = 90 m
s = 240 m - 90 m
The displacement of the ball in the 6.0 seconds is:
s = 150 m to the West
X = -40 m/s.
Y = 5m/s^2 * 6s. = 30 m/s.
A. Tan Ar = Y/X = 30/-40 = -0.750
Ar = -36.9o = Reference angle.
A = -36.9 + 180 = 143.1o CCW = 53.1o W.
of N.
V = X/Cos A = -40/Cos143.1 = 50 m/s @
143.1o CCW or 53.1o W. of N.