A truck travels up a hill with a 12◦
incline.
The truck has a constant speed of 25 m/s.
What is the horizontal component of the
truck’s velocity?
Answer in units of m/s.
consider the extremes.
On a 0° incline, all the speed is along the ramp
on a 90° incline, it's all straight up
Now, which trig function f(x) has
f(0) = 1
f(90°) = 1
?
To find the horizontal component of the truck's velocity, we need to determine the magnitude of the horizontal velocity vector.
The given inclination angle is 12 degrees. We can use this angle to find the horizontal and vertical components of the velocity using trigonometric functions.
1. Identify the horizontal and vertical components:
The horizontal component of the velocity is the magnitude of the velocity vector in the horizontal direction.
The vertical component of the velocity is the magnitude of the velocity vector in the vertical direction.
2. Use trigonometry to find the horizontal component:
The horizontal component of the velocity can be found using the equation:
horizontal component = velocity × cosine(angle)
In this case, the velocity is given as 25 m/s and the angle is 12 degrees. We can plug these values into the equation:
horizontal component = 25 m/s × cosine(12 degrees)
3. Calculate the horizontal component:
Using a calculator or computer, calculate the cosine(12 degrees) and multiply it by the velocity:
horizontal component ≈ 25 m/s × 0.9781
horizontal component ≈ 24.4525 m/s
Therefore, the horizontal component of the truck's velocity is approximately 24.4525 m/s.