Question

Engenders 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 3.5 km/s while moving it through a distance of only 2.0 cm.

a. What acceleration does the gun give this object?
b. Over what time interval does the acceleration take place?

Answer 1

The acceleration given to the object by the gun is 306,250,000 m/s^2, and the time interval over which this acceleration takes place is 11.42 microseconds.

Step-by-step explanation:

Calculating Acceleration and Time Interval

To find the acceleration (a) given to the object by the gun, we can use the kinematic equation v2 = u2 + 2as, where v is the final velocity (3.5 km/s), u is the initial velocity (0 m/s, assuming the object starts from rest), and s is the distance moved (2.0 cm or 0.02 m). Solving for acceleration, we get:

a = (v2 - u2) / 2s

Substituting the given values into the equation, knowing that 3.5 km/s is equal to 3500 m/s, we find:

a = (35002 - 02) / (2 × 0.02)

Therefore, the acceleration is:

a = (12,250,000 m2/s2) / 0.04 m = 306,250,000 m/s2

To calculate the time interval (t) over which this acceleration occurs, we use the equation v = at, which simplifies to t = v/a when starting from rest. Substituting our values, we get:

t = 3500 m/s / 306,250,000 m/s2

The time interval over which the acceleration takes place is thus:

t = 0.00001142 seconds or 11.42 microseconds