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 while moving it through a distance of only 2.5 cm.
(a) What acceleration does the gun give this object?
(b) Over what time interval does the acceleration take place?
Vfinal = sqrt(2aX) = 2700 m/s
X = 0.025 m . Solve for a
Time interval = (distance)/(average speed) = .025m/1350 m/s = ____ s
To answer these questions, we can use the equations of motion. Let's break down each part:
(a) To find the acceleration (a), we can use the equation:
v^2 = u^2 + 2a*s
Where:
- v is the final velocity (2.7 km/s = 2,700 m/s)
- u is the initial velocity (0 m/s, assuming the object starts from rest)
- a is the acceleration (what we're trying to find)
- s is the displacement (2.5 cm = 0.025 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the values:
a = (2,700^2 - 0) / (2 * 0.025)
a ≈ 2,916,000 m/s^2
Therefore, the gun gives the object an acceleration of approximately 2,916,000 m/s^2.
(b) To find the time interval, we can use the equation:
s = ut + (½)at^2
Rearranging the equation, we get:
t = √[(2s) / a]
Substituting the values:
t = √[(2 * 0.025) / 2,916,000]
t ≈ 1.08 × 10^-5 seconds
Therefore, the acceleration takes place over a time interval of approximately 1.08 × 10^-5 seconds.