deceleration?

Question

I know this should be separated into 2 problems in order to solve it. but I've been stuck on where to start, so help. Please?

An inattentive driver is traveling 18.0m/s when he notices a red light ahead. His car is capable of decelerating at a rate of 3.65m/s^2. If it takes him .200s to get the brakes on and he is 20.0m from the intersection when he sees the light, will he be able to stop in time?

Answer 1

You want to find the distance the car has traveled before it comes to a stop. If the distance is less than 20 m than he stopped on time. Otherwise, he didn't.

For .200 s the driver continue to travel at the velocity of 18.0 m/s before he turns on the brake. During this time, he traveled 3.6 m.


We know that v = v_o + at

He stop when the speed is zero therefore

0 = 18 - 3.65 t
t = 4.93

It takes 4.93 seconds after the driver turns on the brake for him to stop.

We know x = x_o + v_o*t + (1/2)at^2

x = (18)(4.93) - (1/2)(3.65)(4.93)^2
x = 44.38 meters

He travel 44.38 meters plus the 3.6 m before he gets the brakes on. He traveled 47.98 m. Nope, he didn't stop in time.

Answer 2

Initial Velocity 'u' = 18 m/s
Deceleration'-a' = 3.65 m/s²
Final velocity 'v' = 0 m/s
Time 't' ................= ?
Here we can use the formula: v = u + at
0 = 18 + (-3.65) x t
0 = 18 - 3.65 t
3.65 t = 18
t = 18 / 3.65 = 4.93 Sec
Time elapsed before applying the brakes= 0.2 sec
Remaining available time = 4.93 - 0.2 = 4.73 s
Dist. travelled before stopping = s = ut + ½at²
s = 18 x 4.73 + ½ (-3.65)(4.73)²
s = 85.14 - ½ x 3.65 x 22.3729
s = 85.14 - 40.831 = 44.309
As the intersection is only 20 m , the driver WILL NOT BE able to stop the car before the red light.
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Answer 3

v^2 = u^2 - 2aS

[S = 20 - 18 x 0.2 = 16.4m; a = - 3.65m/s^2.]

v^2 = 18^2 -[ 2 * 3.65 * 16.4] = 204.28

v = 14.29m/s.

Since his speed is not zero he cannot stop.

Answer 4

S= V0t - 0.5 a t^2