A car starts from rest at a traffic light and accelerates to a velocity of 18 m/s over a distance of 340 m. Which equation would allow you to determine the car's acceleration? (AKS 1c)
Question 8 options:
vf = vi + at
d = vit + 1/2 at2
d = 1/2 (vf + vi) t
vf2 = vi2 + 2ad
The correct equation that would allow you to determine the car's acceleration is:
vf^2 = vi^2 + 2ad
The equation that would allow you to determine the car's acceleration is vf^2 = vi^2 + 2ad.
To determine the car's acceleration, we need to find the equation that relates velocity, distance, and acceleration. The equation that satisfies this condition is:
vf^2 = vi^2 + 2ad
Where:
vf is the final velocity (18 m/s)
vi is the initial velocity (0 m/s since the car starts from rest)
a is the acceleration (unknown)
d is the distance traveled (340 m)
To find the acceleration (a), we rearrange the equation to solve for a:
a = (vf^2 - vi^2) / (2d)
Substituting the known values:
a = (18^2 - 0^2) / (2 * 340)
Simplifying the equation:
a = 324 / 680
Therefore, the car's acceleration can be determined using the equation vf^2 = vi^2 + 2ad.
A subway train travels at 15 m/s. Every 80 seconds it passes a security camera. How far apart are the security cameras? (AKS 1a)
Question 7 options:
5.33 m
80 m
0.19 m
1200 m
To determine the distance between the security cameras, we need to find the equation that relates velocity, time, and distance. The equation that satisfies this condition is:
d = vt
Where:
d is the distance traveled (unknown)
v is the velocity of the subway train (15 m/s)
t is the time it takes to pass a security camera (80 seconds)
Substituting the known values into the equation:
d = 15 m/s * 80 s
Simplifying the equation:
d = 1200 m
Therefore, the distance between the security cameras is 1200 meters.