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It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar had a constant acceleration of -3.9 m/s/s, determine the speed of the Jaguar before it began to skid to a stop.
A.) 48 m/s
B.) -2262 m/s
C.) -48 m/s
D.) 2262 m/s

Respuesta :

kk5996282

Answer:

option c is correct

Explanation:

we know that

2as=vf²-vi²

vf²=2as

vf=√2as

vf=√2*290*(-3.9)

vf=√-2262

vf=-47.6 which is nearly equals to -48 m/s

hope this will helps you

Cricetus

The Speed of the Jaguar will be "-48 m/s".

Side,

  • s = 290 m

Acceleration,

  • a = -3.9 m/s²

As we know,

→ [tex]2as = vf^2-vi^2[/tex]

or,

→ [tex]vf^2 = 2as[/tex]

→ [tex]vf = \sqrt{2as}[/tex]

By substituting the values, we get

→      [tex]= \sqrt{2\times 290\times (-3.9)}[/tex]

→      [tex]= \sqrt{-2262}[/tex]

→      [tex]= -48 \ m/s[/tex]

Thus the above speed i.e., option C) is correct.  

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