A falling object accelerates from
-10.0 m/s to -30.0 m/s. How long does this take?
t= ? s
Remember: falling = -v
lost height = -(Delta) y
g = -9.80 m/s^2
To solve this problem, we can use the equation for acceleration:
acceleration = (change in velocity) / time
Given that the initial velocity (u) is -10.0 m/s, the final velocity (v) is -30.0 m/s, and the acceleration (a) is -9.80 m/s^2 (due to gravity), we can plug in these values into the equation:
-9.80 m/s^2 = (-30.0 m/s - (-10.0 m/s)) / time
Simplifying the equation:
-9.80 m/s^2 = (-30.0 m/s + 10.0 m/s) / time
-9.80 m/s^2 = -20.0 m/s / time
To isolate the time (t):
time = -20.0 m/s / -9.80 m/s^2
Simplifying further:
time = 2.04 s
Therefore, it takes approximately 2.04 seconds for the object to accelerate from -10.0 m/s to -30.0 m/s.
To find the time it takes for the object to accelerate from -10.0 m/s to -30.0 m/s, we can use the equation of motion:
v = u + at
where:
v = final velocity = -30.0 m/s
u = initial velocity = -10.0 m/s
a = acceleration (due to gravity) = -9.80 m/s^2
t = time
Rearranging the equation, we get:
t = (v - u) / a
Substituting the given values:
t = (-30.0 m/s - (-10.0 m/s)) / -9.80 m/s^2
t = (-30.0 m/s + 10.0 m/s) / -9.80 m/s^2
t = -20.0 m/s / -9.80 m/s^2
t ≈ 2.04 seconds
Therefore, it takes approximately 2.04 seconds for the object to accelerate from -10.0 m/s to -30.0 m/s.