Respuesta:
(5±√5) /2 seg.
Explicación:
Velocidad inicial= 10m/s
Aceleración= [tex]- 4m/s^{2}[/tex]
Tiempo= T
Distancia= 10m
----> d= (v)(t) ± (1/2)(a)(t)[tex]{2}[/tex]
10=(10)(T) - (1/2)(4)(T)[tex]{2}[/tex]
10=10T - 2[tex]T^{2}[/tex]
5=5T - [tex]T^{2}[/tex]
[tex]T^{2}[/tex] - 5T + 5 = 0
Formula general: -(-5) ± [tex]\sqrt{(-5)^{2}-4(1)(5) }[/tex] / 2(1)
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Verified answer
Respuesta:
(5±√5) /2 seg.
Explicación:
Velocidad inicial= 10m/s
Aceleración= [tex]- 4m/s^{2}[/tex]
Tiempo= T
Distancia= 10m
----> d= (v)(t) ± (1/2)(a)(t)[tex]{2}[/tex]
10=(10)(T) - (1/2)(4)(T)[tex]{2}[/tex]
10=10T - 2[tex]T^{2}[/tex]
5=5T - [tex]T^{2}[/tex]
[tex]T^{2}[/tex] - 5T + 5 = 0
Formula general: -(-5) ± [tex]\sqrt{(-5)^{2}-4(1)(5) }[/tex] / 2(1)
(5±√5) /2 seg.