21)Slow Sam never wants to hurt anyone so he has only one bullet in his gun which instantly slows down at constant acceleration as it leaves the gun. If the bullet leaves the gun with an initial velocity of 200 m/s and travels 150m before coming to a stop (v=0m/s)
A) what was the bullets avg velocity over the 150 m?
B) what was the bullets acceleration (remember should be neg) over that distance?
A) what was the bullets avg velocity over the 150 m?
B) what was the bullets acceleration (remember should be neg) over that distance?
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You can use one of your three kinematic equations to solve this problem since the bullet has constant acceleration.
A) First, you need to use V=at to find the average velocity of the bullet. Add the 200 to the 0 and divide by two to find the average velocity. (100 m/s)
B) Next, you need to use the equation a = vf-vi/t. This simply means that the acceleration is found by putting the change in velocity over the change in time. That's 200/100, which is two seconds. Therefore, the velocity (200) should be divided by 2, and your answer is -100 m/s/s.
A) First, you need to use V=at to find the average velocity of the bullet. Add the 200 to the 0 and divide by two to find the average velocity. (100 m/s)
B) Next, you need to use the equation a = vf-vi/t. This simply means that the acceleration is found by putting the change in velocity over the change in time. That's 200/100, which is two seconds. Therefore, the velocity (200) should be divided by 2, and your answer is -100 m/s/s.
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A) U=200, V=0, S=150 u = initial velocity, v=velocity, a=acceleration, s= displacement, t= time
200/150 = avg v
B)please use V^2 = u^2 - 2as to find the a.
200/150 = avg v
B)please use V^2 = u^2 - 2as to find the a.