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/s^2,

Determine the speed of the Jaguar before it began to skid.....

Answer 1

Answer: vi = 47.6 m /s

Given

a = -3.90 m/s^2

vf = 0 m/s

d = 290 m

To Find:

vi = ??

Solution

vf^2 = vi^2 + 2*a*d

--> (0 m/s)^2 = vi^2 + 2*(-3.90 m/s^2)*(290 m)

---> 0 m^2/s^2 = vi^2 - 2262 m^2/s^2

2262 m^2/s^2 = vi^2

vi = 47.6 m /s

Answer 2

Answer:

Step-by-step explanation:

We are given: s = 290, u = ?, v = 0 (comes to rest), a = -3.9 and t is irrelevant… using v^2 = u^2 + 2as… substituting gives: 0 = u^2 + 2*(-3.9)*290, Rearranged gives: u = sqrt(2262) which is 47.5604878…. = 47.6 (3s.f.)