A bullet moving with a speed 150 m/s strikes a tree and penetrates 3.5 cm before stopping. Find the magnitude of its acceleration and the time taken to stop.

Vf^2 = Vo^2 + 2ad = 0.

(150)^2 + 2a*3.5 = 0,
22500 + 7a = 0,
a = -3214.3m/s^2.

Vf = Vo + at = 0.
Solve for t and get:
t=(Vf-Vo)/a = (0-150) / -3214.3 = 0.0467s.

Wrong

WRONG ANSWER

To find the magnitude of acceleration, we need to use the equation of motion that relates acceleration (a), initial velocity (u), final velocity (v), and displacement (s):

v^2 = u^2 + 2as

In this case, the initial velocity (u) is 150 m/s, the final velocity (v) is 0 m/s (since the bullet stops), and the displacement (s) is 3.5 cm, which is equivalent to 0.035 m.

Plugging in the values into the equation, we have:

0^2 = (150)^2 + 2a(0.035)

Simplifying the equation further:

0 = 22500 + 0.07a

Rearranging the equation to solve for acceleration (a):

a = -22500 / 0.07

Calculating the value using a calculator, we get:

a ≈ -321,428.57 m/s^2 (rounded to two decimal places)

The negative sign indicates that the acceleration acts in the opposite direction to the initial motion of the bullet.

Now let's find the time taken to stop. We can use the kinematic equation that relates displacement (s), initial velocity (u), acceleration (a), and time (t):

s = ut + (1/2)at^2

In this case, the initial velocity (u) is 150 m/s, the acceleration (a) is approximately -321,428.57 m/s^2, and the displacement (s) is 0.035 m. We need to solve for time (t).

0.035 = 150t + (1/2)(-321,428.57)t^2

Simplifying further:

0 = 150t - 160,714.2857t^2

Rearranging the equation:

160,714.2857t^2 - 150t = 0.035

Solving this quadratic equation, we can use the quadratic formula:

t = [-b ± sqrt(b^2 - 4ac)] / 2a

In our case, a = 160,714.2857, b = -150, and c = -0.035.

Plugging in these values into the formula, we have:

t = [150 ± sqrt((-150)^2 - 4(160,714.2857)(-0.035))] / (2)(160,714.2857)

t = [150 ± sqrt(22500 + 2298.99)] / 321,428.5714

t = [150 ± sqrt(24898.99)] / 321,428.5714

Calculating the values using a calculator, we obtain two solutions:

t ≈ 0.0001718 s (rounded to four decimal places) or t ≈ 0.0048896 s (rounded to four decimal places)

Therefore, the time taken for the bullet to stop is approximately 0.0001718 seconds or 0.0048896 seconds, depending on the direction of rounding.

32.14

You didn't convert cm to m