Starting from position s = 0 m at time t = 0 s, a particle travels on a straight-line path. The particle's velocity is given by the function v = 5 sin (4s2) m/s, where s is in meters and the argument of the sine function is unitless. Find the particle's acceleration when s = 0.25 m and s = 1.25 m.

Question

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Respuesta :

DeniceSandidge DeniceSandidge · Answer 1 · 2019-09-16T07:47:37+02:00

Answer:

acceleration is 0.2181 m/s²

acceleration is 27.05 m/s²

Explanation:

given data

s = 0

time t = 0

velocity V = 5 sin (4s²) m/s

to find out

particle's acceleration

solution

we have given velocity = 5 sin(4s²)

and we know here that velocity = [tex]\frac{ds}{dt}[/tex]

so acceleration will be a = [tex]\frac{dv}{dt}[/tex]

put here velocity v

acceleration = [tex]\frac{dv}{dt}[/tex]

acceleration = [tex]\frac{d(5sin(4s^4))}{dt}[/tex]

acceleration = 5 cos4s² × 8s × [tex]\frac{ds}{dt}[/tex]

acceleration = 5 cos4s² × 8s × 5 sin4s²

acceleration = 200 s ×cos4s² × sin4s²

put here s = 0.25 and s = 1.25

so

acceleration = 200× 0.25 ×cos(4×0.25²) × sin(4×0.25²)

acceleration = 0.2181 m/s²

acceleration = 200× 1.25 ×cos(4×1.25²) × sin(4×1.25²)

acceleration = 27.05 m/s²