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siniylev [52]
3 years ago
11

1) Un cuerpo se mueve desacelerando a razón de 8m/s2, calcula: a) La velocidad que tenía 5 segundos antes de detenerse. b) La di

stancia recorrida los últimos 5 segundos de su trayectoria.
Physics
1 answer:
Galina-37 [17]3 years ago
3 0

Answer:

Initial velocity (u) = 40 m/s

Distance travel in last 5 seconds = 100 m

Explanation:

Given:

Acceleration (a) = 8 m/s²

Final velocity (v) = 0 m/s

Find;

1] Initial velocity before 5 sec

2] Distance travel in last 5 seconds

Computation:

1] Initial velocity before 5 sec

v = u + at

0 = u + (-8)(5)

u - 40 = 0

Initial velocity (u) = 40 m/s

2] Distance travel in last 5 seconds

s = ut + (1/2)(a)(t²)

s = (40)(5) + (1/2)(-8)(5²)

s = 200 - 100

Distance travel in last 5 seconds = 100 m

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P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}

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3 years ago
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