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A bicyclist passing through a city accelerates after he passes the sign post marking the city limits at x=0. His acceleration is constant at 5.0 m/s^2 east. At time t1= 0.0s, he is located at x1 = 6.0m and has a velocity v1=4.0 m/s east.

A) Find his position and velocity at t2= 2.0s.

B) How far is he from the sign post when his velocity is 6.0 m/s.

Respuesta :

MathPhys

Answer:

A. 24 m, 14 m/s

B. 8.0 m

Explanation:

Given:

x₀ = 6.0 m

v₀ = 4.0 m/s

a = 5.0 m/s²

t = 2.0 s

A. Find: x and v

x = x₀ + v₀ t + ½ at²

x = (6.0 m) + (4.0 m/s) (2.0 s) + ½ (5.0 m/s²) (2.0 m/s)²

x = 24 m

v = at + v₀

v = (5.0 m/s²) (2.0 s) + (4.0 m/s)

v = 14 m/s

B. Find x when v = 6.0 m/s.

v² = v₀² + 2a (x − x₀)

(6.0 m/s)² = (4.0 m/s)² + 2 (5.0 m/s²) (x − 6.0 m)

x = 8.0 m

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