Question

It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2 , determine the speed of the Jaguar before it began to skid.​

Answer 1

${\large{\mathfrak{\red{\clubs{\underline{Given :}}}}}}$

It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2.

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${\large{\mathfrak{\blue{\clubs{\underline{To Find :}}}}}}$

What was the speed of Jaguar before he began to skid.

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${\large{\mathfrak{\orange{\clubs{\underline{Solution :}}}}}}$

Here,

${\bf{initial \: velocity = v= 0}}$

${\bf{Speed = 290m}}$

${\bf{Acceleration = -3.9m {s}^{2} }}$

Formula :

${\blue{\bigstar{\pink{\boxed{ {v}^{2} - {u}^{2} = 2as }}}}}$

${\sf{\green{0 - {u}^{2} =2( - 3.9)(290) }}}$

${\sf{\green{ {u}^{2} =2( - 3.9)(290) - 0 }}}$

${\sf{\green{u = \sqrt{2(3.9)(290)} }}}$

After solving :

${\blue{\underbrace{\red{\boxed{\bf{u = 47.6m/s}}}}}}$

So,

The Jaguar was travelling at 47.6m/s before he began to skit.

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${\large{\bf{\underline{\underline{\fcolorbox{green}{red}{\bf{XitzNobitaxX}}}}}}}$

Answer 2

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