A bicycler rider is traveling at 7m/s. During an 8s period, the bicycle rider then slows down with acceleration to a speed of 3 m/s. How far does the bicycle rider travel during the 8s?

d=7m/s
s=3m/s
t=8s

t x s = d/t x t
d = s x t

d = (7 m/s + 3 m/s) / 2 * 8 s

d = 10 m/s / 2 * 8 s

d = 5 m/s * 8 s

d = 40 m

The bicycle rider travels a distance of 40 meters during the 8 second period.

To calculate the distance traveled by the bicycler rider during the 8-second period, we can use the formula:

Distance (d) = Speed (s) x Time (t)

In this case, the initial speed (when the rider starts slowing down) is 7 m/s (meters per second), and the final speed (when the rider reaches a speed of 3 m/s) is 3 m/s. The time period is 8 seconds.

Plugging in the values into the formula:

d = 3 m/s x 8 s
d = 24 meters

Therefore, the bicycle rider travels a distance of 24 meters during the 8-second period.

Vo = 7 m/s

V = 3 m/s
t = 8 s.

V = Vo + a*t
Solve for a. It should be negative.

d = Vo*t + 0.5a*t^2
Solve for d.