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 4.4 km/s while moving it through a distance of 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 acceleration in part (a) of the question, we can use the formula for acceleration:
acceleration (a) = change in velocity (Δv) / change in time (Δt)
However, the information provided in the question doesn't directly give us the change in velocity or the change in time. We are given the initial velocity of the object (0 km/s), the final velocity (4.4 km/s), and the distance the object traveled (1.6 cm).
To find the change in velocity, we can subtract the initial velocity from the final velocity:
Δv = 4.4 km/s - 0 km/s = 4.4 km/s
To find the change in time, we need to convert the distance traveled from centimeters to kilometers, since velocity is given in km/s:
distance = 1.6 cm = 0.016 m
Now, we can use the formula for average velocity:
average velocity = distance / time
Since the average velocity is equal to the final velocity (4.4 km/s), we can rearrange the formula to solve for time:
time = distance / average velocity = 0.016 m / 4.4 km/s = 0.000016 km / 4.4 km/s
Now, we have the change in velocity (Δv = 4.4 km/s) and the change in time (Δt = 0.000016 s). We can substitute these values into the formula for acceleration to find the answer:
acceleration (a) = Δv / Δt = 4.4 km/s / 0.000016 s
To convert the units, we need to ensure that the units for velocity and time are consistent. Let's convert the velocity from km/s to m/s:
acceleration (a) = (4.4 km/s * 1000 m/km) / 0.000016 s
Simplifying the equation:
acceleration (a) = 275,000 m/s / 0.000016 s
Finally, we can calculate the acceleration:
a ≈ 17,187,500,000 m/s²
Therefore, the gun gives the object an acceleration of approximately 17,187,500,000 m/s².
Now, let's move on to part (b) of the question, which asks for the time interval over which the acceleration takes place. We have already calculated the change in time as Δt = 0.000016 s, which means the acceleration takes place over this time interval.