When a car is traveling at 21 m/s on level ground, the stopping distance is found to be 18 m. This distance is measured by pushing hard on the brakes so that the wheels skid on the pavement. The same car is driving at the same speed down an incline that makes an angle of 7.6° with the horizontal direction. What is the stopping distance now, as measured along the incline?
To find the stopping distance along the incline, we need to break down the motion into horizontal and vertical components.
First, let's calculate the horizontal component of the car's velocity. Since the car is moving at 21 m/s, the horizontal velocity remains constant regardless of the incline. Therefore, the horizontal velocity is still 21 m/s.
Next, we need to find the vertical component of the car's velocity. Since the incline makes an angle of 7.6° with the horizontal, we can calculate the vertical velocity using trigonometry. The vertical component is given by Vvertical = Vtotal * sin(theta), where Vtotal is the total velocity and theta is the angle of the incline.
Vvertical = 21 m/s * sin(7.6°)
Vvertical = 21 m/s * 0.1317
Vvertical ≈ 2.76 m/s
Now, we can analyze the motion along the incline. The vertical component of velocity will cause the car to accelerate downward due to gravity, while the horizontal component remains constant.
Since the car is being braked, it will come to a stop when the vertical displacement matches the initial vertical velocity and acceleration due to gravity (negative value since it is acting downward). Therefore, we can use the equation: V^2 = V0^2 + 2aΔy, where V is the final velocity, V0 is the initial velocity, a is the acceleration, and Δy is the vertical displacement.
The final vertical velocity (V) is 0 m/s (since the car stops), the initial vertical velocity (V0) is 2.76 m/s, and the acceleration (a) is -9.8 m/s^2 (acceleration due to gravity). We can rearrange the equation to solve for Δy:
0^2 = (2.76 m/s)^2 + 2 * (-9.8 m/s^2) * Δy
0 = 7.6176 m^2/s^2 - 19.6 m/s^2 * Δy
19.6 m/s^2 * Δy = 7.6176 m^2/s^2
Δy = 7.6176 m^2/s^2 / 19.6 m/s^2
Δy ≈ 0.3889 m
Finally, the stopping distance along the incline is the horizontal displacement (distance traveled horizontally) when the car comes to a stop. Since the horizontal velocity is constant, we can use the equation: distance = velocity * time.
The time it takes for the car to come to a stop along the incline is the same as the time it takes for the car to drop vertically by Δy. We can use the equation Δy = 0.5 * a * t^2 to find the time (t).
0.3889 m = 0.5 * (-9.8 m/s^2) * t^2
0.3889 m = -4.9 m/s^2 * t^2
t^2 = 0.3889 m / -4.9 m/s^2
t^2 ≈ -0.0793 s^2
Since time cannot be negative, we can ignore the negative sign. Solving for t gives:
t ≈ √(0.0793 s^2)
t ≈ 0.2814 s
Now, we can calculate the stopping distance along the incline:
distance = velocity * time
distance = 21 m/s * 0.2814 s
distance ≈ 5.925 m
Therefore, the stopping distance along the incline is approximately 5.925 meters.