A particle travels horizontally between two
parallel walls separated by 18.4 m. It moves
toward the opposing wall at a constant rate
of 5.8 m/s. Also, it has an acceleration in the
direction parallel to the walls of 1.4 m/s2 .
Is there a question here?
lol. He forgot to ask a question
To find the time it takes for the particle to reach the opposing wall, we can use the equation of motion:
s = ut + 1/2at^2
Where:
s = distance traveled (18.4 m)
u = initial velocity (5.8 m/s)
a = acceleration (1.4 m/s^2)
t = time
Rearranging the equation, we get:
t = (-u ± √(u^2 + 2as)) / a
Substituting the values into the equation:
t = (-5.8 ± √(5.8^2 + 2 * 1.4 * 18.4)) / 1.4
Simplifying further:
t = (-5.8 ± √(33.64 + 51.52)) / 1.4
t = (-5.8 ± √85.16) / 1.4
Now, we have two possible solutions for time, since we used the ± sign:
t₁ = (-5.8 + √85.16) / 1.4
t₂ = (-5.8 - √85.16) / 1.4
Calculating the values:
t₁ = ( -5.8 + 9.23 ) / 1.4
t₁ = 3.43 / 1.4
t₁ ≈ 2.45 seconds
t₂ = ( -5.8 - 9.23 ) / 1.4
t₂ = -15.03 / 1.4
t₂ ≈ -10.74 seconds
Since time cannot be negative in this context, the only valid solution is t₁ ≈ 2.45 seconds. Therefore, it takes approximately 2.45 seconds for the particle to reach the opposing wall.