Answer:
1. 3 m
2. 27 s
Explanation:
1. "A car traveling at +33 m/s sees a red light and has to stop. If the driver can accelerate at -5.5 m/s², how far does it travel?"
Given:
v₀ = 33 m/s
v = 0 m/s
a = -5.5 m/s²
Unknown: Δx
To determine the equation you need, look for which variable you don't have and aren't solving for. In this case, we aren't given time and aren't solving for time. So look for an equation that doesn't have t in it.
Equation: v² = v₀² + 2aΔx
Substitute and solve:
(0 m/s)² = (33 m/s)² + 2(-5.5 m/s²) Δx
Δx = 3 m
2. "A plane starting from rest at one end of a runway accelerates at 4.8 m/s² for 1800 m. How long did it take to accelerate?"
Given:
v₀ = 0 m/s
a = 4.8 m/s²
Δx = 1800 m
Unknown: t
Equation: Δx = v₀ t + ½ a t²
Substitute and solve:
1800 m = (0 m/s) t + ½ (4.8 m/s²) t²
t ≈ 27 s
<span><span>anonymous </span> 4 years ago</span>Any time you are mixing distance and acceleration a good equation to use is <span>ΔY=<span>V<span>iy</span></span>t+1/2a<span>t2</span></span> I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890.
For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time.
Add the two times together for the total.
The alternative is to calculate the initial and final velocity so that you have more information to work with.
