A truck travels up a hill with a 13◦
incline.
The truck has a constant speed of 18 m/s.
What is the horizontal component of the
truck’s velocity?
Answer in units of m/s
X = 18 * Cos13.
Well, if the truck is going up a hill at a 13◦ incline, it's going to need all the help it can get. So, the horizontal component of its velocity is actually going to be 18 m/s. It's trying to defy gravity here, so we can't blame it for going full speed ahead!
To find the horizontal component of the truck's velocity, we need to use trigonometry and the given information.
The truck's velocity can be represented as a vector with both horizontal and vertical components. The vertical component is the speed at which the truck is moving up the hill, while the horizontal component is the speed in the horizontal direction.
Given:
Incline angle = 13°
Speed of the truck = 18 m/s
We can use the trigonometric formula:
Horizontal component = Speed * cos(angle)
Plugging in the values:
Horizontal component = 18 m/s * cos(13°)
Using a calculator, we can find the cosine of 13°:
cos(13°) ≈ 0.9781
Now, we can calculate the horizontal component:
Horizontal component ≈ 18 m/s * 0.9781
Horizontal component ≈ 17.6 m/s
Therefore, the horizontal component of the truck's velocity is approximately 17.6 m/s.
To determine the horizontal component of the truck's velocity, we need to find the component of its velocity that is parallel to the ground (in the horizontal direction).
To do this, we can use trigonometry. The horizontal component of velocity can be found using the following formula:
Horizontal component of velocity = Total velocity * cosine(angle of incline)
In this case, the total velocity of the truck is given as 18 m/s and the angle of incline is given as 13°.
Using the formula, we substitute the values and calculate the horizontal component of velocity:
Horizontal component of velocity = 18 m/s * cos(13°)
Calculating this expression, we get:
Horizontal component of velocity ≈ 18 m/s * 0.9744
Simplifying further:
Horizontal component of velocity ≈ 17.538 m/s
Therefore, the horizontal component of the truck's velocity is approximately 17.538 m/s.