A skateboarder initially traveling at 1.8 m/s accelerates at a rate of 1.5 m/s squared for 2.5 s. How far does the skateboarder go in this time?
What is the equations used, variable solving for, and of course the solution.
idk bruh
To find the distance traveled by the skateboarder, we can use the kinematic equation:
\[ d = v_0t + \frac{1}{2}at^2 \]
where:
d = distance traveled
v₀ = initial velocity
t = time
a = acceleration
Given:
v₀ = 1.8 m/s
a = 1.5 m/s²
t = 2.5 s
We need to plug in these values into the equation to calculate the distance.
Substituting the given values:
\[ d = (1.8 m/s)(2.5 s) + \frac{1}{2}(1.5 m/s²)(2.5 s)^2 \]
Simplifying the equation:
\[ d = 4.5 m + \frac{1}{2}(1.5 m/s²)(6.25 s^2) \]
\[ d = 4.5 m + 4.6875 m \]
\[ d = 9.1875 m \]
Therefore, the skateboarder travels a distance of 9.1875 meters in 2.5 seconds.
To solve this problem, you can use the equation of motion that relates distance (d), initial velocity (u), acceleration (a), and time (t):
d = u*t + (1/2)*a*t^2
In this case, the initial velocity (u) is given as 1.8 m/s, the acceleration (a) is given as 1.5 m/s^2, and the time (t) is given as 2.5 s.
Plugging these values into the equation, we have:
d = (1.8 m/s) * (2.5 s) + (1/2)*(1.5 m/s^2)*(2.5 s)^2
Simplifying this expression further:
d = 4.5 m + 0.5*(1.5 m/s^2)*(6.25 s^2)
d = 4.5 m + 0.5*9.375 m
d = 4.5 m + 4.6875 m
d = 9.1875 m
So, the skateboarder goes approximately 9.19 meters in this time.