Will the Car Stop in Time to Avoid Hitting the Child in the School Zone?

[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...

 

[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.

 

[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

 

[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