A bullet is fired through a board 8.0 cm thick in such a way that the bullet\'s
ID: 1532976 • Letter: A
Question
A bullet is fired through a board 8.0 cm thick in such a way that the bullet's line of motion is perpendicular to the face of the board. The initial speed of the bullet is 650 m/s and it emerges from the other side of the board with a speed of 550 m/s. Find the acceleration of the bullet as it passes through the board. Find the total time the bullet is contact with the board. A speedboat moving at 36.0 m/s approaches a no-wake marker 300 m abroad. The shows the boat with a constant acceleration of -3.80 m/s^2 by reducing the throttle. How long does it take the boat to reach the busy? What is the velocity of the boat when it reaches the busy?Explanation / Answer
a) we can use the equation vf^2=v0^2+2ad
where vf is final speed = 550 m/s
v0=initial speed = 650 m/s
a=accel (to be found)
d= distance traveled = 8cm = 0.08 m
so:
550^2=650^2+2a(0.08) => a =750000 m/s/s
b) acceleration = change in speed/time
time = change in speed/acceleration = (-100m/s)/-750,000 m/s/s
time = 1.33x10^-4 secs
2.
First working formula,
S = VT - (1/2)aT^2
where
S = distance of boat from buoy = 100 m
V = initial velocity of boat = 36 m/sec
T = time for boat to reach the buoy
a = acceleration = - 3.8 m/sec^2
Substituting values,
100 = 36T - (1/2)(3.8)T^2
100 = 36T - 1.9T^2
Rewriting the above,
1.9T^2 - 36T + 100 = 0
Using the quadratic formula,
T = 3.38 sec.
Next working formula is
Vf - Vo = aT
where
Vf = velocity of boat as it reaches the buoy
and all the other terms have been defined.
Substituting values,
Vf - 36 = (-3.8)(3.38)
Vf = 23.156 m/sec.
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