Question

a bicyclist approaches the credt of a hill at 4.5 m/s. she accelerates down the hill at a rate of 0.40 m/s^2 for 12 s. how far does she move down the hill during this time interval?​

Answer 1

25.3m

Step-by-step explanation:

Given parameters:

Final velocity = 4.5m/s

Acceleration = 0.4m/s²

Time taken = 12s

Unknown:

Distance she moved down hill = ?

Solution:

To solve this problem, we use the expression below:

v² = u² + 2aS

v is the final velocity

u is the initial velocity

a is the acceleration

S is the distance

4.5² = 0² + (2 x 0.4 x S)

20.25 = 0.8S

S = 25.3m