Engineers are developing new types of guns that might someday be used to launch satellites as if they were bullets. One such gun can give a small object a velocity of 2.9 km/s, moving it through only 1.6 cm.
(a) What acceleration does the gun give this object?
(b) Over what time interval does the acceleration take place?
To find the answers to these questions, we'll need to use the equations of motion.
First, let's find the acceleration of the object.
We can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity = 2.9 km/s
u = initial velocity = 0 (since the object starts from rest)
s = displacement = 1.6 cm = 0.016 m
a = acceleration (what we're trying to find)
Plugging in the known values into the equation, we get:
(2.9 km/s)^2 = (0 m/s)^2 + 2a(0.016 m)
Simplifying the equation, we have:
8.41 km^2/s^2 = 0.032a m
To convert km^2/s^2 to m^2/s^2, we need to multiply by 1,000,000:
8410 m^2/s^2 = 0.032a m
Now, we can solve for a:
a = 8410 m^2/s^2 / 0.032 m
a = 262812.5 m/s^2
So, the gun gives the object an acceleration of 262812.5 m/s^2.
Next, let's find the time interval over which the acceleration takes place.
We can use another equation of motion to find the time:
v = u + at
We already know v (2.9 km/s), u (0 m/s), and a (262812.5 m/s^2). Plugging these values into the equation:
2.9 km/s = 0 m/s + 262812.5 m/s^2 * t
Simplifying the equation, we have:
2.9 km/s = 262812.5 m/s^2 * t
To solve for t, we can rearrange the equation:
t = (2.9 km/s) / (262812.5 m/s^2)
t = 0.01104 s
Therefore, the acceleration takes place over a time interval of 0.01104 seconds.