Question

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

Answer 1

v=0
u=?
s=290m
a=(-3.90m/s^2)


v^2-u^2=2as
(0)^2-u^2=2×(-39/10)×290
-u^2= -2262
u=47.560m/s

Answer 2

The speed of the Jaguar before it began to skid is 47.56 m/s.

Explanation:

Given that,

Distance, d = 290 m

Acceleration of the jaguar, $a=-3.9\ m/s^2$

Finally it stops, v = 0

Let u is the speed of the Jaguar before it began to skid. It is given by using third equation of kinematics as :

$v^2-u^2=2ad$

$-u^2=2ad$

$-u^2=2\times (-3.9)\times 290$

$u^2=2\times (3.9)\times 290$

u = 47.56 m/s

So, the speed of the Jaguar before it began to skid is 47.56 m/s. Hence, this is the required solution.

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Equation of motion

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