# Kinematics for car in 2D

• Mitchtwitchita
In summary, the conversation discusses a problem where a car traveling at 40 km/h needs to stop before hitting a child who runs onto the road 13 m ahead of it. By using equations and calculations, it is determined that the car will stop in time before hitting the child.

## Homework Statement

The speed limit in a school zone is 40 km/h (about 25 mi/h). A driver sees a child run onto the road 13 m ahead of his car. He applies the breaks, and the car decelerates at a uniform rate of 8.0 m/s^2. If the driver's reaction time is 0.25 s, will the car stop before hitting the child?

## Homework Equations

d = Vot + 1/2at^2

## The Attempt at a Solution

40 km/h x [1000 m/1 km] x [1 h/3600 s] = 11 m/s

Vo = 11 m/s
X = ?
Xo = 0
a = -8.0 m/s^2
t = 0.25 s

X - Xo = Vot + 1/2at^2
X = (11 m/s)(0.25 s) + 1/2(-8.0 m/s^2)(0.25 s)^2
= 2.5 m

Vo = 11 m/s
V = 0
X = ?
Xo = 2.5 m
a = -8.0 m/s^2

V^2 = Vo^2 + 2a(X - Xo)
X = Xo + [V^2 - Vo^2/2a]
= 2.5 + [0 - (11 m/s)^2/2(-8.0 m/s^2)
= 10 m

Therefore, 2.5 m + 10 m = 12.5 m, and the car will stop in time.

Seems right, can anybody please verify this outcome?

Dunno if I'm right, but this is how I see this problem.

First, I calculate the distance car has traveled in that 0.25 seconds

$$d=v*t=\frac{100m}{9s}*\frac{1s}{4}=\frac{25m}{9}$$

After that, the car slows down

$$v^{2}=v_{0}^{2}+2ad$$
$$0=\frac{100}{9}^{2}-2*8d$$
$$d=\frac{100*100}{81*2*8}$$

Now to sum up those 2 distances

$$d=\frac{25}{9}+\frac{10000}{1296}=\frac{13600}{1296}\approx10.5m$$

But once again, I don't know if I'm rigth (Although I agree the boy will live )

Oh, I see where I went wrong. No deceleration during the first part, and a constant velocity. Thank-you for your help!

You're welcome :D