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 .
Displacement s = (u+v)*t/2 (t refers to delta time)
= (0.45 + 2.7)*6/2
= 3.15*3
= 9.45 m
Below are the choices that can be found elsewhere:
A. (4.9 × 10-14 newtons) · tan(30°)
<span>B. (4.9 × 10-14 newtons) · sin(30°) </span>
<span>C. (4.9 × 10-14 newtons) · cos(30°) </span>
<span>D. (4.9 × 10-14 newtons) · arctan(30°) </span>
<span>E. (4.9 × 10-14 newtons) · arccos(30°)
</span>
<span>Force is proportional to the angle made by the velocity with respect to the magnetic field. It is maximum when velocity is perpendicular to the magnetic field and minimum when the velocity is parallel to the magnetic field. It is proportional to sin of the angle. In this problem it will be proportional to sin(30)</span>
The distance covered by the acorn is 3.136 m.
<u>Explanation:</u>
The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.
Then using the second law of equation,
Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so
So the distance covered by the acorn is 3.136 m.
Answer:
v(t)= (d/dt)x(t)
Explanation:
The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of x with respect to t. Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point t
0 is the rate of change of the position function, which is the slope of the position function
x
(
t
)
at t
0
.
