Please help me learn how to do this problem, step by step please.
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 4.1 km/s while moving it through a distance of only 1.5 cm.
(a) What acceleration does the gun give this object? (b) Over what time interval does the acceleration take place?
You know displacement and final velocity of an object starting from rest.
Use the relevant mechanics equations:
s = (at^2)/2
v = at
Where: s=0.015[m] and v=4100[m/s]
Rearranging the velocity equation to give an expression for time taken (t).
Use this to substitute for t into the displacement equation, and rearrange to obtain an expression for acceleration.
Use given values in this expression to calculate the acceleration (the answer to part a).
Substitute back into the rearranged velocity equation to calculate the time interval (the answer to part b)
To solve this problem, we can use the kinematic equations of motion. The first step is to identify the given values and what we need to find.
Given:
- Initial velocity (u) = 0 m/s (since the object starts from rest)
- Final velocity (v) = 4.1 km/s = 4100 m/s
- Displacement (s) = 0.015 m (1.5 cm)
(a) To find acceleration (a), we can use the equation:
v^2 = u^2 + 2as
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (4100^2 - 0^2) / (2 * 0.015)
Now, we can calculate the acceleration using a calculator:
a ≈ 4,946,000 m/s^2
So, the acceleration the gun gives to the object is approximately 4,946,000 m/s^2.
(b) To find the time interval, we can use the equation:
v = u + at
Rearranging the equation, we get:
t = (v - u) / a
Substituting the given values:
t = (4100 - 0) / 4,946,000
Now, we can calculate the time interval using a calculator:
t ≈ 0.00083 seconds
So, the acceleration takes place over approximately 0.00083 seconds.
Thus, the answers to the problem are:
(a) The gun gives the object an acceleration of approximately 4,946,000 m/s^2.
(b) The acceleration takes place over approximately 0.00083 seconds.