What would be the magnitude of the electric field 0.75 m from a 0.63 C master charge and what would be the force on a 0.50 C tes
1 answer:
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Answer:
The total distance is 381.5 [m]
Explanation:
In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0
a = acceleration = 2 [m/s²]
t = time = 7 [s]
Vf = 0 + (2*7)
Vf = 14 [m/s]
With this velocity, we can calculate the displacement using the following expression.
where
x = distance traveled [m]
14² = 0 + (2*7*x)
x = 196/(14)
x = 14 [m]
Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.
The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.
We must calculate the final velocity.
Vf = final velocity [m/s]
Vi = initial velocity = 14 [m/s]
a = desacceleration = 4 [m/s²]
t = time = 8 [s]
Vf = 24 + (4*8)
Vf = 56 [m/s]
With this velocity, we can calculate the displacement using the following expression.
where
x = distance traveled [m]
56² = 14² + (2*4*x)
x = 2940/(8)
x = 367.5 [m]
Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.
Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].