CAN SOMEONE HELP ME PLEASE!? After chasing its prey, a cougar leaves skid marks that are 236 m in length. Assuming the cougar skidded to a complete stop with a constant acceleration of -2.87 m/s^2, identify the speed of the cougar before it began to skid.

u=36.8m/s

Explanation:

because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations

u^2=v^2-2ā*s. where:

u^2 stands for intial velocity

v^2 stands for final velocity

since the cougar skidded to a complete stop the final velocity is zero.

u^2=v^2-2ā*s

u^2=(0)^2 -2(-2.87 m/s^2)*236 m

u^2=0+5.74m/s^2* 236m

u^2=1354.64m^2/s^2

u=√1354.64m^2/s^2

u=36.8m/s (approximate value)

when ever the acceleration is constant you can use one of the following equation to find the required value.

1. v = u + at.(no s)

2. s= 1/2(u+v)t.(no ā)

3. s=ut + 1/2at^2.( no v)

4. v^2=u^2 + 2āS.(no t).5. s= vt - 1/2at^2.(no u)

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000