Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .
Question
What was the initial momentum of the bullet before collision?
Answer:
10 Kg.m/s
Explanation:
Momentum is a product of velocity of an object in m/s and its mass in kgs hence numerically expressed as p=mv where p is momentum, v is velocity and m is mass. Substituting m for 0.2 kg and v for 50 m/s then p=0.2*50=10 kg.m/s
It's by number of protons