You are driving at 20 m/s when you notice a tree blocking the road, 30 meters ahead of you. It takes you half a second to react before slamming on the brakes. Your car decelerates at a constant 10 m/s2. How far does your car travel before stopping? Will you hit the tree?
everything happens starting at t=.5
a = -10
v = -10t + c
when t = .5 , v = 20
20 = -10(.5) + c
c = 25
v = -10t + 25 , for t ≥ .5
s = -5t^2 + 25t + k
when t = .5 , the car goes goes another 20(.5) or 10 m
so when t = .5, the remaining distance is 20 m
20 = -5(.5)^2 + 25(.5) + k
k = 8.75
s = -5t^2 + 25t + 8.75
the car stops when v = 0
0 = -10t + 25
t = 2.5
when t = 2.5
s = -5(2.5)^2 + 25(2.5) + 8.75
= 40
Unfortunately the tree was in the way, and yes, you are going to hit it
To find out how far your car travels before stopping, we can use the kinematic equation:
\(d = v_i * t + \frac{1}{2} * a * t^2\)
Where:
d = distance
\(v_i\) = initial velocity
a = acceleration
t = time
Given:
\(v_i = 20 \, \text{m/s}\)
\(a = -10 \, \text{m/s}^2\) (negative because deceleration)
\(t = 0.5 \, \text{s}\)
Plugging these values into the equation, we can calculate the distance:
\(d = 20 \, \text{m/s} * 0.5 \, \text{s} + \frac{1}{2} * (-10 \, \text{m/s}^2) * (0.5 \, \text{s})^2\)
Simplifying the equation:
\(d = 10 \, \text{m} - 1.25 \, \text{m}\)
\(d = 8.75 \, \text{m}\)
Therefore, your car will travel 8.75 meters before coming to a stop.
Now, let's check whether you will hit the tree. The tree is 30 meters ahead, and your car stops after traveling 8.75 meters. Since 8.75 meters is less than 30 meters, you will not hit the tree.