A particle travels horizontally between two parallel walls separated by 18.4 m. It moves toward the opposing wall at a constant rate of 9.4 m/s. Also, it has an acceleration in the direction parallel to the walls of 5.8 m/s^2
What will be its speed when it hits the
opposing wall?
Answer in units of m/s
It will take the particle t = 18.4/9.4 = 1.957 s to go from wall to wall. In doing so, it will acquire a velocity component parallel to the wall equal to
Vy = 5.8*t = 11.35 m/s
Compute the vector sum of 11.35 m/s and 9.4 m/s at right angles
To find the speed of the particle when it hits the opposing wall, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (speed when it hits the wall)
u = initial velocity (given as 9.4 m/s)
a = acceleration (given as 5.8 m/s^2)
s = distance traveled (given as 18.4 m)
First, let's calculate the displacement (s) covered by the particle using the formula:
s = ut + 0.5at^2
Since the particle is moving horizontally, the displacement (s) will be equal to the distance between the walls:
s = 18.4 m
Now we can rearrange the formula to solve for time (t):
18.4 = (9.4)t + 0.5(5.8)(t^2)
This equation is a quadratic equation in terms of time (t). Solving it will give us the value of time (t).
Once we calculate the value of time (t), we can substitute it back into the equation of motion to find the final velocity (v):
v^2 = (9.4)^2 + 2(5.8)(18.4)
Now, take the square root of v^2 to find the final velocity (v) when it hits the opposing wall.
v = √(9.4^2 + 2(5.8)(18.4))
By substituting the given values into the equation and evaluating it, we can find the final velocity (v) in units of m/s.