A motor cyclist travelling at 12m/s decelerate at3m/s(squared) how far does he travel while coming to rest?

Use V=Vo+at to find time. Vo is the original velocity and V will be 0 because you are trying to see how far the motor cyclist has gone when he stops.

Find time, then use x=Xo+Vot+(1/2)at^2 to find distance.

Yes, Emily's approach will give the correct answer.

You can also use another kinematics relation to solve for distance x:
V1^2-V0^2=2ax
where
V1=final velocity = 0
V0=initial velocity = 12 m/s
a=acceleration = -3 m/s²
x=distance travelled in m.

To find the distance traveled while the motorcyclist comes to a stop, you can use the kinematic equation:

v^2 = u^2 + 2as

Where:
v = final velocity (in this case, 0 m/s, as the motorcyclist comes to rest)
u = initial velocity (12 m/s)
a = deceleration (-3 m/s^2)
s = distance traveled

Let's substitute the given values into the equation:

0^2 = 12^2 + 2(-3)s

Simplifying further:

0 = 144 - 6s

Rearranging the equation to solve for s:

6s = 144

s = 144/6

s = 24 meters

Therefore, the motorcyclist will travel 24 meters while coming to rest.