Starting from rest, a car accelerates at 2.5 m/s^2 up a hill that is inclined 5.4 ^\circ above the horizontal.
how far the car travel horizontally in 15s?
I'm assuming the angle of inclination is 5.4 deg.
a(h) = 2.5m/s^2 * cos5.4 = 2.49m/s^2 =
hor. component of acceleration.
d = Vo*t + 0.5at^2,
d = 0 + 0.5*2.49*(15)^2 = 280m.
Well, if the car is going up a hill, it seems like it's really trying to reach new heights! But let's calculate the horizontal distance it travels in 15 seconds.
To find the horizontal distance, we need to break down the acceleration into its horizontal and vertical components.
The vertical component of the acceleration can be found using the formula: A_vertical = A * sin(theta), where A is the total acceleration and theta is the angle of inclination.
A_vertical = 2.5 m/s^2 * sin(5.4°)
Now, we need to find the time it takes for the car to reach a velocity of 15 m/s. We can use the formula: V = A * t, where V is the final velocity, A is the acceleration, and t is the time.
15 m/s = 2.5 m/s^2 * t
t = 15 m/s / 2.5 m/s^2
t = 6 seconds
Now that we have the vertical acceleration and the time it takes to reach the final velocity, we can calculate the vertical distance traveled using the formula: d = 0.5 * A_vertical * t^2.
d_vertical = 0.5 * (2.5 m/s^2 * sin(5.4°)) * (6 s)^2
Finally, we can calculate the horizontal distance using the formula: d_horizontal = V * t - 0.5 * A_horizontal * t^2, where A_horizontal = A * cos(theta).
A_horizontal = 2.5 m/s^2 * cos(5.4°)
d_horizontal = 15 m/s * 6 s - 0.5 * (2.5 m/s^2 * cos(5.4°)) * (6 s)^2
And there you have it! Just plug in the values and you'll find the answer to your question.
To find the horizontal distance the car has traveled, we need to consider the car's acceleration and time.
First, let's break down the given information:
- Acceleration (a) = 2.5 m/s^2
- Inclination angle (θ) = 5.4 degrees
- Time (t) = 15 seconds
To calculate the horizontal distance (d), we can use the equation:
d = v₀ * t + (1/2) * a * t^2
Where:
- v₀ = initial velocity (which is 0 m/s since the car starts from rest)
- t = time
- a = acceleration
Now, let's calculate the horizontal distance traveled by substituting the given values into the equation:
d = (0) * t + (1/2) * 2.5 * t^2
d = 0 + 1.25 * t^2
Since we are given a time of 15 seconds:
d = 1.25 * (15)^2
d = 1.25 * 225
d = 281.25 meters
Therefore, the car travels horizontally for a distance of 281.25 meters in 15 seconds.
To find the horizontal distance traveled by the car, we need to break down the motion into its horizontal and vertical components.
First, let's find the vertical (vertical) component of the acceleration. Since the car is moving up the hill, the vertical component of the acceleration will be given by:
acceleration_vertical = acceleration * sin(angle)
Here, acceleration = 2.5 m/s^2 and the angle = 5.4 degrees.
So, acceleration_vertical = 2.5 * sin(5.4)
Next, let's find the time taken for the car to travel 15 seconds. Since the motion is only in the vertical direction, we will use the equation:
time_vertical = time
Given in the problem, time = 15 seconds.
Now, let's find the vertical displacement using the equation of motion:
displacement_vertical = (initial_velocity * time_vertical) + (0.5 * acceleration_vertical * time_vertical^2)
Here, the car starts from rest, so initial_velocity = 0.
So, displacement_vertical = 0 + (0.5 * acceleration_vertical * time_vertical^2)
Finally, let's find the horizontal displacement using the equation:
displacement_horizontal = displacement_vertical / cos(angle)
Here, displacement_vertical is the vertical displacement that we just calculated, and angle = 5.4 degrees.
By substituting the values into the equation, we can find the horizontal displacement.