A spaceship moving with an initial velocity of 58.0 meters/second experiences a uniform acceleration and attains a final velocity of 153 meters/second. What distance has the spaceship covered after 12.0 seconds?
d = V0 * t + 1/2 a t ** 2
where
average acceleration a = (V1 - V0) / dt
V0 = 58 m/s
V1 = 153 m/s
dt = 12 s
a = (153-58)/12 m/(s*s)
a = 7.9 m/(s*s)
d = 58 * 12 m/s *s + .5 * 7.9 * (12 * 12) m/(s*s) * (s*s)
d = 696 + 1140 m
d = 1836 m
P.S.
I forgot to multiply by the .5 in the second to last step .
It should read :
d = 696 + .5 * 1140 m
d = 1266 m
100
lad
To find the distance covered by the spaceship, we can use the equation of motion:
Distance = (Initial Velocity × Time) + (1/2) × (Acceleration × Time^2)
Given:
Initial Velocity (u) = 58.0 meters/second
Final Velocity (v) = 153 meters/second
Time (t) = 12.0 seconds
To find the acceleration (a) of the spaceship, we can use the following formula:
v^2 = u^2 + 2a × d
Rearranging this equation, we get:
a = (v^2 - u^2) / (2 × d)
Let's start by finding the acceleration using the values of initial and final velocities:
a = (153^2 - 58.0^2) / (2 × d)
Solving this equation, we can find the acceleration (a).
Next, once we have the value of acceleration (a), we can substitute it into the equation of motion to find the distance covered (d):
Distance = (Initial Velocity × Time) + (1/2) × (Acceleration × Time^2)
Substitute the given values to find the distance covered by the spaceship after 12.0 seconds.