Truck accelerating down a ramp

[Dern]

Homework Statement

An 18 wheeler going down a road with a 10 degree downgrade is traveling at 25 m/s and accelerating at 1.7 m/s^2. When the truck has descended an additional 20 meters of vertical distance (if air is neglected), what could he expect his velocity to reach if the incline stays the same?

Relevant Equations

I know you can solve the problem by finding the distance traveled down the ramp that corresponds to a 20 m vertical drop (x = 20/sin(10) = 115.2 m). Then you can use the kinematic formula: v(f)^2 = v(i)^2 + 2as to get an answer of 32 m/s.

However, I also wanted to solve this problem by focusing on the vertical drop info to find the time required to do so, and then substitute that time into the equation: y = v(i,y)t + 1/2a(y)t^2.

After tilting the axes to fit the 10 degree down ramp, I first found the tilted y distance drop that corresponds with the 20 m vertical drop: y = -20cos(10) = -19.7 m. Then I used the formula, y = v(i,y)t + 1/2a(y)t^2, to get:

-19.7 = 0*t + 1/2(-9.8cos(10))t^2

since the initial velocity is only in the tilted x direction. Solving for t, I arrived at t = 2.02 s. However, when I substitute this time into the equation, v(f) = v(i) + at, I get a final velocity of 28.4 m/s. What am I doing wrong?

 

[TSny]

After tilting the axes to fit the 10 degree down ramp, I first found the tilted y distance drop that corresponds with the 20 m vertical drop: y = -20cos(10) = -19.7 m. Then I used the formula, y = v(i,y)t + 1/2a(y)t^2, to get:

-19.7 = 0*t + 1/2(-9.8cos(10))t^2

since the initial velocity is only in the tilted x direction.

Welcome to PF!

In your tilted x-y system, does the y-coordinate of the truck change as it goes down the ramp?

Since the acceleration of the truck is in the direction parallel to the ramp, does the truck have any y-component of acceleration?

 

[AlexB23]

Homework Statement: An 18 wheeler going down a road with a 10 degree downgrade is traveling at 25 m/s and accelerating at 1.7 m/s^2. When the truck has descended an additional 20 meters of vertical distance (if air is neglected), what could he expect his velocity to reach if the incline stays the same?
Relevant Equations: I know you can solve the problem by finding the distance traveled down the ramp that corresponds to a 20 m vertical drop (x = 20/sin(10) = 115.2 m). Then you can use the kinematic formula: v(f)^2 = v(i)^2 + 2as to get an answer of 32 m/s.

However, I also wanted to solve this problem by focusing on the vertical drop info to find the time required to do so, and then substitute that time into the equation: y = v(i,y)t + 1/2a(y)t^2.

After tilting the axes to fit the 10 degree down ramp, I first found the tilted y distance drop that corresponds with the 20 m vertical drop: y = -20cos(10) = -19.7 m. Then I used the formula, y = v(i,y)t + 1/2a(y)t^2, to get:

-19.7 = 0*t + 1/2(-9.8cos(10))t^2

since the initial velocity is only in the tilted x direction. Solving for t, I arrived at t = 2.02 s. However, when I substitute this time into the equation, v(f) = v(i) + at, I get a final velocity of 28.4 m/s. What am I doing wrong?

I am not a homework helper, but I remember doing a problem like this.

Remember, 10 degrees can be converted into rise over run. So, you will have to use trig to calculate the rise over run. You are given the rise: 20 meters, and an angle. Note that the angle between the shortest side (20 meter rise) and the horizontal run is 90°. So, now, you must calculate the hypotenuse.


You will be using SAA (side angle angle).

 

[Steve4Physics]

After tilting the axes to fit the 10 degree down ramp, I first found the tilted y distance drop that corresponds with the 20 m vertical drop: y = -20cos(10) = -19.7 m. Then I used the formula, y = v(i,y)t + 1/2a(y)t^2, to get:

-19.7 = 0*t + 1/2(-9.8cos(10))t^2

You have used a component of free-fall acceleration (g= 9.8m/s^2) - but nothing is free-falling! You are told the truck's acceleration is 1.7m/s^2 in the direction of the incline. So g is irrelevant in this particular problem.

 

[PeroK]

You have used a component of free-fall acceleration (g= 9.8m/s^2) - but nothing is free-falling! You are told the truck's acceleration is 1.7m/s^2 in the direction of the incline. So g is irrelevant in this particular problem.

Note that $9.8\sin(10°) = 1.7$.

 

[Steve4Physics]

Note that $9.8\sin(10°) = 1.7$.

If I understand the OP's intention, I think that after the rotation, the truck is travelling on a horizontal road with a horizontal acceleration of $1.7m/s^2$.

The acceleration in the $10^o$-from-the vertical-direction is then $1.7\sin(10°)$ and not $9.8\sin(10°)$. Or have I misunderstood?

 

[erobz]

If I understand the OP's intention, I think that after the rotation, the truck is travelling on a horizontal road with a horizontal acceleration of $1.7m/s^2$.

The acceleration in the $10^o$-from-the vertical-direction is then $1.7\sin(10°)$ and not $9.8\sin(10°)$. Or have I misunderstood?

I think you have misunderstood. I think it's just a kinematics equation not to be troubling the OP with forces. They have concealed the motive force for the acceleration of a truck on a ramp. Sure $g$ is involved in reality, but it wasn't in my mind either until @PeroK calculated the intentionally hidden stuff!

 

[PeroK]

If I understand the OP's intention, I think that after the rotation, the truck is travelling on a horizontal road with a horizontal acceleration of $1.7m/s^2$.

The problem with the OP's approach is that once we consider only the motion along the incline, we lose sight of the vertical drop that was one of the variables in the equation.

The acceleration in the $10^o$-from-the vertical-direction is then $1.7\sin(10°)$ and not $9.8\sin(10°)$. Or have I misunderstood?

You've misunderstood. They gave you $1.7m/s^2$ in the question, but it turns out that it is the expected acceleration due to gravity down the incline in any case.

 

[Steve4Physics]

I think you have misunderstood. I think it's just a kinematics equation not to be troubling the OP with forces. They have concealed the motive force for the acceleration of a truck on a ramp. Sure $g$ is involved in reality, but it wasn't in my mind either until @PeroK calculated the intentionally hidden stuff!

I viewed the problem entirely kinematically and never considered or mentioned forces!

I only referred to $g$ because the OP (incorrectly) used it - I was arguing that it should not be used!

 

[erobz]

I viewed the problem entirely kinematically and never considered or mentioned forces!

I only referred to $g$ because the OP (incorrectly) used it - I was arguing that it should not be used!

Oh. I've misunderstood the misunderstanding!

 

[Steve4Physics]

The problem with the OP's approach is that once we consider only the motion along the incline, we lose sight of the vertical drop that was one of the variables in the equation.

You've misunderstood. They gave you $1.7m/s^2$ in the question, but it turns out that it is the expected acceleration due to gravity down the incline in any case.

I think the OP's attempt to analyse the problem with the $10^{\circ}$ rotation was just out of 'experimental interest' and I was just trying to expain why it was wrong.

It doesn't matter really that $1.7m/s^2$ is the 'free' acceleration down the incline.(Though well done for spotting it!) The given acceleration is arbitrary; for example if the question had given the acceleration to be, say, $2.0m/s^2$ then the solution and my explanation would still be applicable. Just my opinion though!