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 2.7 km/s, moving it through only 1.5 cm.
(a) What acceleration does the gun give this object?
(b) Over what time interval does the acceleration take place?
vf=at or
a= vf/time for b)
for a) vf^2=2*acceleration*distance
(2700)^2=2*acceleration*.015 solve for acceleration
To determine the acceleration and time interval for the given scenario, we can use the equations of motion.
(a) Acceleration (a) can be calculated using the equation:
v^2 = u^2 + 2as
where:
v = final velocity (2.7 km/s)
u = initial velocity (0 m/s, as the object starts from rest)
a = acceleration (to be calculated)
s = displacement (1.5 cm = 0.015 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the values, we get:
a = (2.7 km/s)^2 / (2 * 0.015 m)
Converting 2.7 km/s to m/s, we have:
a = (2700 m/s)^2 / (2 * 0.015 m)
Calculating this, we find:
a ≈ 2,160,000 m/s^2
Therefore, the gun gives the object an acceleration of approximately 2,160,000 m/s^2.
(b) The time interval can be calculated using the equation:
v = u + at
where all the variables have the same meaning as before.
Rearranging the equation, we have:
t = (v - u) / a
Plugging in the values, we get:
t = (2.7 km/s - 0 m/s) / 2,160,000 m/s^2
Converting 2.7 km/s to m/s, we have:
t = (2700 m/s - 0 m/s) / 2,160,000 m/s^2
Calculating this, we find:
t ≈ 1.25 * 10^(-3) seconds
Therefore, the acceleration takes place over approximately 1.25 * 10^(-3) seconds.