# Car deceleration physics problem

1. Nov 1, 2005

### rafael_josem

Hi, I need help with the following problem....

The speed limit of a school zone is 40km/h. A Driver driving at this speed sees a child that crosses the street 13m in front of the car. He applies the brakes and desacelerates at 8m/s^2. If the reaction time of the driver is 0.50s, will the car stop before it hits the child?

I did this:

40km/h : 11.11m/s

X = v^2 - V0^2/2a
X = 0 - 11.11^2/-16
X = 7.7m <== He drives 7.7m before the car stops.

Now I look for the time that the car will last to stop.

X = (v + v0/2)t
7.7 = (0+11.11/2)t
t = 1.4s

Now I add 0.5s + 1.4s = 1.9s

X = (v+v0/2)t
X = (v + 11.11/2)1.9
X = 10.5 M

According to the book, it's wrong...

Thanks...

2. Nov 1, 2005

### FredGarvin

Try rearanging your method. You have solved for two different values that physically represent the same thing, i.e. the stopping distance.

Try calculating the time to stop first by using
$$V = V_o + a t$$

That will give you your 1.4 seconds to decelerate from the given speed to a stop.

Now take that time (plus distance during the reaction time) and plug and chug with:
$$\Delta X = V_o t + \frac{1}{2} a t^2$$

See what you come up with then.

Last edited: Nov 2, 2005
3. Nov 2, 2005

### verty

To find the stopping distance, I would say:

t_1 = 0.5s (reaction time)
X = v_0*t_1 + (v^2 - v_0^2)/(2a)

You could also do this in two steps:
X_1 = v_0*t_1
X_2 = (v^2 - v_0^2)/(2a)
X_t = X_1 + X_2

4. Nov 2, 2005

### Ehsan Tarkeh

X=Vt >> X=11.11*0.5 = 5.55 which means when the driver put his leg on the brake the distance between the car and the student is 7.45 m and not 13 meter

V^2=2ax >> x= (11.11^2)/(2*8) = 7.71 which means that the car will move 7.71 meter to stop, while the distance between the car and a child is less than this amount (7.45<7.71) so the car will hit the child

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