Starting from rest, a car accelerates at 3.0 m/s^2 up a hill that is inclined 5.9 ^\circ above the horizontal.
1-How far horizontally has the car traveled in 14 s?
2-How far vertically has the car traveled in 14 s?
distance up the hill: 1/2 *3.0*t^2
horizontally: distanceuphill*cosTheta
vertically: distanceuphill*sinTheta
Vo= 0m/s
Vox = Voy = 0m/s > Xo = Xy = 0
a= 3 m/s^2
x is adjacent so Ax = 3cos(5.9)
Ay = 3sin(5.9)
Use X=Xo+Vox+1/2Axt^2 to get X1. Remember, Xo = 0...
To solve these questions, we first need to calculate the vertical and horizontal components of the car's acceleration.
1. Horizontal distance traveled in 14 s:
Since the car starts from rest, the initial velocity is zero. Therefore, we can use the formula:
Distance = (Initial velocity x time) + (0.5 x acceleration x time^2)
The initial velocity is zero, and the time is 14 s. To find the horizontal distance traveled, we need to consider the horizontal component of acceleration. The horizontal component of the acceleration is given by:
horizontal acceleration = acceleration x cos(theta)
where theta is the angle of inclination of the hill, which is 5.9 degrees.
Using the formula for distance, with the appropriate values substituted:
Distance = (0 x 14) + (0.5 x [3.0 x cos(5.9)]) x 14^2 = (0.5 x 3.0 x cos(5.9) x 14^2)
Now, we can calculate this distance:
Distance = 0.5 x 3.0 x cos(5.9) x 14^2
2. Vertical distance traveled in 14 s:
Similarly, using the vertical component of acceleration, which is given by:
vertical acceleration = acceleration x sin(theta)
Using the formula for distance again:
Distance = (0 x 14) + (0.5 x [3.0 x sin(5.9)]) x 14^2 = (0.5 x 3.0 x sin(5.9) x 14^2)
Now, we can calculate this distance as well:
Distance = 0.5 x 3.0 x sin(5.9) x 14^2
To find the horizontal distance traveled by the car, we need to find the car's horizontal velocity at the end of the 14-second time interval.
1. Calculate the horizontal velocity:
To find the horizontal component of the car's velocity, we use the formula:
v_horizontal = v_initial * cos(angle), where v_initial is the initial velocity and angle is the incline angle.
Since the car starts from rest (initial velocity, v_initial = 0), the horizontal velocity will also be zero initially. Thus, the horizontal velocity remains constant throughout the motion.
v_horizontal = 0 * cos(5.9°)
v_horizontal = 0
2. Calculate the horizontal distance:
To find the horizontal distance, we use the formula:
distance = velocity * time
distance = v_horizontal * time
distance = 0 * 14
distance = 0
Therefore, the horizontal distance traveled by the car in 14 seconds is 0 meters.
To find the vertical distance traveled by the car, we need to find the car's vertical displacement at the end of the 14-second time interval.
1. Calculate the vertical displacement:
To find the vertical component of the car's displacement, we use the formula:
d_vertical = (v_initial * sin(angle) * time) + (0.5 * acceleration * time^2),
where v_initial is the initial velocity, angle is the incline angle, acceleration is the car's acceleration, and time is the given time interval.
Since the car starts from rest (initial velocity, v_initial = 0), the first term simplifies to 0. Thus, the vertical displacement reduces to:
d_vertical = 0.5 * acceleration * time^2
d_vertical = 0.5 * 3.0 m/s^2 * (14 s)^2
d_vertical = 0.5 * 3.0 * 196
d_vertical = 294 m
Therefore, the vertical distance traveled by the car in 14 seconds is 294 meters.