the solid fuel rocket boosters used to launch the space shuttle are able to lift the shuttle 45 kilometers above earths surface. During the 2.00 min that the boosters operate, the shuttle accelerates from rest to a speed of nearly
7.50 x 10^2 m/s. What is the magnitude of the shuttle's acceleration?
2.51 m/s^2
To find the magnitude of the shuttle's acceleration, we can use the following equation:
acceleration (a) = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = 0 m/s (the shuttle starts from rest)
Final velocity (v) = 7.50 x 10^2 m/s
Time (t) = 2.00 min = 2.00 x 60 s = 120 s
Plugging in the values:
a = (v - u) / t
= (7.50 x 10^2 m/s - 0) / 120 s
= (7.50 x 10^2 m/s) / 120 s
Calculating this expression:
a = (7.50 x 10^2) / 120
= 6.25 m/s^2
Therefore, the magnitude of the shuttle's acceleration is 6.25 m/s^2.
To find the magnitude of the shuttle's acceleration, we can use the formula:
a = (v - u) / t
where:
a = acceleration
v = final velocity
u = initial velocity
t = time
In this case, the initial velocity (u) is 0 m/s because the shuttle starts from rest. The final velocity (v) is 7.50 x 10^2 m/s, and the time (t) is 2.00 minutes.
First, we need to convert the time to seconds because the formula requires time in seconds. There are 60 seconds in one minute, so:
t = 2.00 min * 60 s/min = 120 s
Now we can substitute the values into the formula:
a = (7.50 x 10^2 m/s - 0 m/s) / 120 s
Simplifying this equation:
a = (7.50 x 10^2 m/s) / 120 s
To evaluate this, we divide 7.50 x 10^2 by 120:
a ≈ 6.25 m/s²
So, the magnitude of the shuttle's acceleration is approximately 6.25 m/s².