A motor cyclist travelling at 12m/s decelerates at 3m/s square

a) how long does he take to come to a rest
b) how far does he travel in coming to rest

average speed during stop = 12/2 = 6

v = Vi - a t
0 = 12 - 3 t to stop (get v = 0)
so
t = 12/3 = 4 seconds to stop

distance = average speed * time
= 6 * 4
= 24 meters

a) The motorcyclist is decelerating at a rate of 3 m/s². To find the time it takes for him to come to a rest, we can use the equation v² = u² + 2as, where v is the final velocity (0 m/s in this case), u is the initial velocity (12 m/s), a is the acceleration (-3 m/s²), and s is the distance traveled.

Plugging in the values, we get 0² = 12² + 2(-3)s. Simplifying, we have 0 = 144 - 6s. Rearranging, we find 6s = 144. Dividing by 6, we get s = 24.

So, it takes the motorcyclist 24 seconds to come to a rest.

b) To find the distance the motorcyclist travels in coming to rest, we can use the equation v² = u² + 2as again.

Plugging in the values, we get 0² = 12² + 2(-3)s. Simplifying, we have 0 = 144 - 6s. Rearranging, we find 6s = 144. Dividing by 6, we get s = 24.

Therefore, the motorcyclist travels a distance of 24 meters in coming to rest.

To solve these questions, we can use the following equations of motion:

1) v = u + at - Equation 1
2) s = ut + (1/2)at^2 - Equation 2

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

a) To find the time taken to come to rest, rearrange Equation 1 to solve for t:

v = u + at
0 = 12 + (-3t)
-12 = -3t
t = 4 seconds

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

b) To find the distance traveled, rearrange Equation 2 to solve for s:

s = ut + (1/2)at^2
s = 12(4) + (1/2)(-3)(4)^2
s = 48 - 24
s = 24 meters

Therefore, the motorcyclist travels a distance of 24 meters in coming to rest.

To answer these questions, we can use the equations of motion for uniformly decelerated motion.

a) To find how long the motorcyclist takes to come to a rest, we can use the equation:

v = u + at

where:
v = final velocity (0 m/s, since the motorcycle comes to a rest)
u = initial velocity (12 m/s)
a = acceleration (deceleration, -3 m/s²)
t = time

Rearranging the equation to solve for time (t), we get:

t = (v - u) / a

Substituting the given values, we can calculate the answer:

t = (0 - 12) / (-3)
t = 4 seconds

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

b) To determine how far the motorcyclist travels while decelerating, we can use the equation:

s = ut + (1/2)at²

where:
s = distance
u = initial velocity (12 m/s)
t = time (4 s, as calculated in part a)
a = acceleration (deceleration, -3 m/s²)

Plugging in the values:

s = (12)(4) + (1/2)(-3)(4)²
s = 48 - 24
s = 24 meters

Therefore, the motorcyclist travels a distance of 24 meters while coming to a rest.