an auto, moving too fast on a horizontal stretch of mountain road, slides off the road, falling into deep snow 43.9 meters below the road and 87.7 meters beyond the edge of the road.
a. How long did the auto take to fall?
b. How fast was it going when it left the road?
To find the answers, we can use the equations of motion.
a. Since the car falls vertically downward, we can use the equation h = (0.5)gt^2, where h is the vertical distance fallen and g is the acceleration due to gravity (approximately 9.8 m/s^2). Rearranging the equation, we have t = sqrt(2h/g).
Given that the vertical distance fallen (h) is 43.9 meters, we can calculate the time (t) it took for the car to fall:
t = sqrt(2 * 43.9 / 9.8)
t = sqrt(8.918)
t ≈ 2.988 seconds
Therefore, the auto took approximately 2.988 seconds to fall.
b. Since the car slid horizontally, we can use the equation d = vt, where d is the horizontal distance traveled and v is the initial velocity. Rearranging the equation, we have v = d/t.
Given that the horizontal distance slid (d) is 87.7 meters and the time (t) is 2.988 seconds, we can calculate the initial velocity (v):
v = 87.7 / 2.988
v ≈ 29.33 m/s
Therefore, the auto was going approximately 29.33 m/s when it left the road.
To calculate the time it took for the auto to fall and the speed at which it left the road, we can use the equations of motion and consider the vertical and horizontal components of the motion separately.
a. To find the time it took for the auto to fall, we can use the equation for vertical motion:
Δy = V_y0 * t + (1/2) * a * t^2
Where:
Δy is the vertical displacement (43.9 meters)
V_y0 is the initial vertical velocity (which is zero as the auto is sliding off the road horizontally)
a is the acceleration due to gravity (-9.8 m/s^2, considering it as negative)
Rearranging the equation, we get:
43.9 = 0 * t + (1/2) * (-9.8) * t^2
Simplifying further:
-4.9t^2 = 43.9
Dividing both sides by -4.9:
t^2 = -43.9 / -4.9
t^2 = 9
Taking the square root of both sides, we get:
t ≈ 3 seconds
Therefore, it took approximately 3 seconds for the auto to fall.
b. To find the speed at which the auto left the road, we can use the equation for horizontal motion:
Δx = V_x0 * t
where:
Δx is the horizontal displacement (87.7 meters)
V_x0 is the initial horizontal velocity (which is the speed we want to find)
t is the time (3 seconds)
Rearranging the equation, we get:
V_x0 = Δx / t
Substituting the given values:
V_x0 = 87.7 / 3
V_x0 ≈ 29.23 meters per second
Therefore, the auto was going approximately 29.23 meters per second when it left the road.
Oh my, the same question with different numbers again. See below:
http://www.jiskha.com/display.cgi?id=1386888436#1386888436.1386888818