A runaway truck lane heads uphill at 30◦ to the horizontal.
If an out of control 15000 kg truck enters the lane going at 30 m/s , how far along the ramp does it go? The acceleration due to gravity is 10 m/s2.
To calculate how far the truck goes along the ramp, we need to determine the horizontal and vertical components of its initial velocity.
Given:
Mass of the truck (m) = 15000 kg
Initial velocity (v) = 30 m/s
Angle of the ramp (θ) = 30 degrees
Acceleration due to gravity (g) = 10 m/s^2
First, we will find the horizontal and vertical components of the initial velocity. The horizontal component (v_x) is calculated using the formula:
v_x = v * cos(θ)
v_x = 30 * cos(30)
v_x ≈ 26.0 m/s
The vertical component (v_y) is calculated using the formula:
v_y = v * sin(θ)
v_y = 30 * sin(30)
v_y ≈ 15.0 m/s
Now, we need to calculate the time it takes for the truck to reach the top of the hill. Since there is no horizontal acceleration, the time is the same for both horizontal and vertical motion.
The formula for time (t) is:
t = v_y / g
t = 15.0 / 10
t = 1.5 seconds
Now, we can calculate the distance traveled along the ramp (d) using the horizontal component of the initial velocity and the time taken:
d = v_x * t
d = 26.0 * 1.5
d = 39.0 meters
Therefore, the truck goes approximately 39.0 meters along the ramp.
To find the distance the truck goes along the ramp, we can break down the problem into two components: the horizontal and vertical components of motion.
Let's start by calculating the vertical motion of the truck.
Given:
The angle of the runaway truck lane with respect to the horizontal is 30°.
The acceleration due to gravity is 10 m/s^2.
We can find the vertical component of the truck's velocity initially and at the end when it stops moving vertically.
Vertical component of initial velocity, v₀y = v₀ * sin(θ)
where v₀ is the initial velocity (30 m/s) and θ is the angle of the ramp (30°).
v₀y = 30 * sin(30°) = 30 * 0.5 = 15 m/s
Final vertical velocity, vfy = 0 m/s (the truck comes to rest vertically).
Using the kinematic equation for vertical motion:
vfy^2 = v₀y^2 + 2 * a * Δy
where vfy is the final vertical velocity, v₀y is the initial vertical velocity, a is the acceleration (gravity: -10 m/s^2), and Δy is the vertical displacement.
0 = (15 m/s)^2 + 2 * (-10 m/s^2) * Δy
Solving for Δy, we get:
Δy = [(15 m/s)^2] / [2 * (10 m/s^2)] = 11.25 m
Now that we know the vertical displacement is 11.25 m, we can calculate the horizontal distance traveled by the truck.
The horizontal distance, Δx, can be found using the formula:
Δx = v₀x * t
where v₀x is the horizontal component of the initial velocity, which is given by v₀x = v₀ * cos(θ), and t is the time taken to travel the distance Δy vertically.
v₀x = 30 * cos(30°) = 30 * 0.866 = 25.98 m/s
To find the time, t, let's use the equation:
Δy = v₀y * t + (1/2) * a * t^2
Plugging in the known values,
11.25 = (15 m/s) * t + (1/2) * (-10 m/s^2) * t^2
Rearranging and solving the quadratic equation, we get:
5t^2 - 15t - 11.25 = 0
Using the quadratic formula, t = [-(-15) ± √((-15)^2 - 4 * 5 * -11.25)] / (2 * 5)
t = [15 ± √(225 + 225)] / 10
t ≈ [15 ± √450] / 10
t ≈ [15 ± 21.21] / 10
Taking the positive value, t ≈ (15 + 21.21) / 10 ≈ 3.12 s
Now, we can calculate the horizontal distance traveled:
Δx = v₀x * t
Δx = 25.98 m/s * 3.12 s ≈ 81.14 m
Therefore, the truck travels approximately 81.14 meters along the runway ramp.