To find the average acceleration of the bullet as it penetrates the sandbag, we can use the following formula for acceleration:
[a = \frac{v_f - v_i}{t}]
Where:
- (a\) is the acceleration
- (v_f\) is the final velocity
- (v_i\) is the initial velocity
- (t\) is the time taken
In this case, the bullet is moving horizontally, so its initial vertical velocity ((v_{iy})) is 0 m/s because there's no initial vertical motion. The bullet penetrates a distance of 5 cm, which we need to convert to meters for consistent units. 1 cm = 0.01 m, so 5 cm = 0.05 m.
Now, we need to find the time taken (\(t\)) for the bullet to penetrate this distance. To do this, we can use the following kinematic equation:
[d = v_i t + \frac{1}{2} a t^2]
Where:
- (d) is the distance (0.05 m)
- (v_i) is the initial velocity (in the horizontal direction, which is 436 m/s)
- (a) is the acceleration (which we want to find)
- (t) is the time taken
Since the bullet is moving horizontally, there is no acceleration in the horizontal direction (assuming no external forces act on it). Therefore, \(a\) in the horizontal direction is 0 m/s². This means that the only force acting on the bullet in the horizontal direction is its initial velocity, and it will maintain this velocity as it penetrates the sandbag.
Now, we can use the simplified kinematic equation to find (t):
[0.05 m = (436 m/s) t]
Solve for (t):
[t = \frac{0.05 m}{436 m/s} \approx 0.0001147 s]
Now that we have the time, we can calculate the acceleration:
[a = \frac{v_f - v_i}{t}]
Since the horizontal velocity ((v_i) remains constant and is 436 m/s, and there's no change in velocity ((v_f)), the acceleration ((a)) is indeed 0 m/s².
So, the average acceleration of the bullet as it penetrates the sandbag is 0 m/s² in the horizontal direction. This means the bullet maintains a constant horizontal velocity throughout its penetration.