A motor cyclist travelling at 12m/s decelerates at 3m/s^2.

(a)How long does he take to come to rest?
(b)How far does he travel in coming to rest?

a. vf=vi+at

0=12-3t solve for t in seconds
b. distance= avgvelocity*time=6m/s*time

9m/s

To find the answers to these questions, we need to use the equations of motion. There are three main equations of motion that we can use, but we will focus on the equation:

v = u + at

Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration,
and t is the time taken.

Let's solve each part of the question step by step:

(a) How long does the motorcyclist take to come to rest?

Given:
Initial velocity, u = 12 m/s
Acceleration, a = -3 m/s^2 (negative sign indicates deceleration)
Final velocity, v = 0 m/s (since the motorcyclist comes to rest)

Using the equation v = u + at, we can rearrange it to solve for time (t):

0 = 12 + (-3)t

Rearranging the equation further:

3t = 12

Dividing both sides by 3:

t = 4 seconds

Therefore, it takes the motorcyclist 4 seconds to come to rest.

(b) How far does the motorcyclist travel in coming to rest?

To find the distance traveled, we can use another equation of motion:

s = ut + 0.5at^2

Where:
s is the distance traveled

Given:
Initial velocity, u = 12 m/s
Acceleration, a = -3 m/s^2
Time, t = 4 seconds

Substituting the given values into the equation, we can calculate the distance traveled:

s = (12)(4) + 0.5(-3)(4)^2
s = 48 - 0.5(16)(16)
s = 48 - 128
s = -80 meters

The negative sign indicates that the distance traveled is in the opposite direction of motion.

Therefore, the motorcyclist traveled a distance of 80 meters in coming to rest.