superman flies to a meteor traveling at 440km/hr.superman slows it down at a rate of -24.8m/s^2.how long before the man of steel stops the meteor?how far did he travel with the meteor?
To find the time it takes for Superman to stop the meteor, we can use the equation of motion:
v = u + at
where:
v = final velocity (0 m/s since the meteor stops)
u = initial velocity (440 km/hr convert to m/s)
a = acceleration (-24.8 m/s^2)
t = time (unknown)
First, let's convert the initial velocity from kilometers per hour (km/hr) to meters per second (m/s).
1 km/hr = (1000 m) / (3600 s) = 0.2778 m/s
So, the initial velocity of the meteor is:
u = 440 km/hr × 0.2778 m/s = 122.22 m/s
Using the equation of motion, we have:
0 = 122.22 m/s + (-24.8 m/s^2) • t
Rearranging the equation to solve for time, we get:
t = -122.22 m/s / (-24.8 m/s^2)
t ≈ 4.93 seconds
Therefore, it takes approximately 4.93 seconds for Superman to stop the meteor.
Next, let's calculate the distance Superman travels with the meteor using the equation of motion:
s = ut + (1/2) a t^2
where:
s = distance (unknown)
u = initial velocity (122.22 m/s)
a = acceleration (-24.8 m/s^2)
t = time (4.93 seconds)
Plugging in the values, we get:
s = (122.22 m/s)(4.93 s) + (1/2)(-24.8 m/s^2)(4.93 s)^2
Simplifying the equation, we find:
s ≈ 301.95 meters
Therefore, Superman travels approximately 301.95 meters with the meteor before he stops it.