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.
The answer is 90 meters. Henry don't know.
Yes. The answer is indeed 90 meters.
it 90 meters y'all don't know what you are talking about
Its actually 11.7 if youre on Quest from UT
it be 90 meters yo. Getcho facts straight yo
To solve this problem, we can use the principles of Newtonian mechanics. Let's break it down step by step:
1. First, let's find the component of the truck's weight (force due to gravity) that acts along the direction of the ramp. The weight force is given by the formula: F = m * g, where m is the mass of the truck (15000 kg) and g is the acceleration due to gravity (10 m/s²). So, the weight force along the ramp can be calculated as: F = 15000 kg * 10 m/s² = 150000 N.
2. Now, let's find the component of gravity force that opposes the motion of the truck along the ramp. Since the truck is going uphill, the component of gravity force that opposes the motion is given by: F(opposing) = F * sin(30°). So, the opposing force is: F(opposing) = 150000 N * sin(30°) ≈ 75000 N.
3. Next, we can use Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the sum of the opposing force and any other forces acting on the truck. Since the truck is out of control, we can ignore any other forces, and the net force is just the opposing force. So, we have: F(net) = F(opposing) = m * a, where a is the acceleration of the truck along the ramp.
4. Now, let's rearrange the formula to solve for the acceleration: a = F(opposing) / m. Plugging in the values, we get: a = 75000 N / 15000 kg ≈ 5 m/s².
5. Finally, we can use the kinematic equation, s = ut + (1/2)at², to find the distance (s) the truck travels along the ramp. Here, u is the initial velocity (30 m/s), t is the time taken, and a is the acceleration. Since we want to find the distance traveled, we can set the final velocity (v) to be zero because the truck comes to a stop. So, the formula becomes: s = ut + (1/2)at² ≈ 30 m/s * t + (1/2) * 5 m/s² * t².
Given that the truck comes to a stop, we can find the time taken by setting the final velocity to zero and solving for t in the equation: v = u + at. Plugging in the values, we have: 0 = 30 m/s + (-5 m/s²) * t, which gives us t ≈ 6 seconds.
Plugging this value of t back into the distance formula, we get: s = 30 m/s * 6 s + (1/2) * 5 m/s² * (6 s)² = 180 m + 90 m = 270 m.
Therefore, the truck travels approximately 270 meters along the ramp.
Yes, the answer is 90 meters because h=(Vo^2-V^2)/g. g stands for gravity, which is 10 m/s^2. I don't know where Henry got the two from, but it doesn't go there. Anyways... h=(900-0)/10, which = 90 meters. Sorry Henry. Much luck to you in your future in physics.
all these answers are wrong
Correctin:
Vo = 30m/s @ 30o.
Yo = 30*sin30 = 15 m/s.
h = (Y^2-Yo^2)/2g.
h = (0-225)/-20 = 11.25 m.
h = (V^2-Vo^2)/2g.
h = (0-900)/-20 = 45 m.