You're driving down the highway at 20m/s when you see a snowman you don't want to crush 35 m in front of you. Your reaction time before stepping on the brakes is 0.5s, and acceleration created by your brakes is 10 m/s^2 what was the distance travelled before coming to a stop?
I would appreciate it if you show me the equation you're going to use.
d = Vo*t + (V^2-Vo^2)/2a
d = 20*0.5 + (0-20^2)/-10 = 50 m.
Note: Vo*t is the distance traveled
during the reaction time.
The 2nd Eq was derived from
V^2 = Vo^2 + 2a*d. Solve for d.
To solve this problem, we will utilize the following equation:
\[ \text{{distance}} = \text{{initial velocity}} \times \text{{time}} + \frac{1}{2} \times \text{{acceleration}} \times \text{{time}}^2 \]
Let's break down the given information into variables:
Initial velocity (vi) = 20 m/s
Reaction time (t1) = 0.5 s
Acceleration (a) = -10 m/s^2 (negative sign indicates deceleration)
Final velocity (vf) = 0 m/s (because the car comes to a stop)
Distance to the snowman (d) = 35 m
To find the distance traveled before coming to a stop, we will calculate the distance traveled during the reaction time and the distance traveled during the braking process. We can break down the problem into two separate steps:
Step 1: Calculate the distance traveled during the reaction time.
Using the equation: distance = initial velocity × time, the distance during the reaction time is:
\[ d1 = vi \times t1 \]
Step 2: Calculate the distance traveled during the braking process.
Using the equation: distance = initial velocity × time + 0.5 × acceleration × time^2, the distance during the braking process is:
\[ d2 = vi \times t2 + \frac{1}{2} \times a \times t2^2 \]
Here, the initial velocity for the braking process is 0 because the car starts decelerating.
Now, let's calculate each step individually.
Step 1:
\[ d1 = vi \times t1 = 20 \, \text{m/s} \times 0.5 \, \text{s} = 10 \, \text{m} \]
Step 2:
\[ d2 = vi \times t2 + \frac{1}{2} \times a \times t2^2 \]
To find t2, we need to consider the velocity change from the reaction time. We can use the equation:
\[ vf = vi + a \times t \]
Since vf is 0 m/s (car comes to a stop), we have:
\[ vf = vi + a \times t2 \]
\[ 0 = 20 \, \text{m/s} + (-10 \, \text{m/s}^2) \times t2 \]
\[ t2 = \frac{-20 \, \text{m/s}}{-10 \, \text{m/s}^2} = 2 \, \text{s} \]
Now, we can calculate d2:
\[ d2 = vi \times t2 + \frac{1}{2} \times a \times t2^2 \]
\[ d2 = 0 \times 2 + \frac{1}{2} \times -10 \times 2^2 \]
\[ d2 = 0 - 20 \, \text{m} \]
Finally, the total distance traveled before coming to a stop is the sum of d1 and d2:
\[ \text{Total distance} = d1 + d2 \]
\[ \text{Total distance} = 10 \, \text{m} + (-20 \, \text{m}) \]
\[ \text{Total distance} = -10 \, \text{m} \]
Hence, the distance traveled before coming to a stop is -10 meters.
To find the distance traveled before coming to a stop, we can use the equation of motion with constant acceleration:
d = v₀t + (1/2)at²
where:
d = distance traveled
v₀ = initial velocity
t = time
a = acceleration
Given data:
v₀ = 20 m/s (initial velocity)
t = 0.5 s (reaction time)
a = -10 m/s^2 (deceleration due to braking)
Now, let's break down the problem into two parts:
1. Reaction Time:
During the reaction time, the car is still moving at a constant velocity. Therefore, the distance traveled during this time can be calculated using the equation:
d₁ = v₀t
Substituting the values, we have:
d₁ = 20 m/s * 0.5 s
d₁ = 10 m
2. Braking Time:
After the reaction time, the car starts decelerating due to the brakes. In this case, we need to calculate the distance traveled during this deceleration period using the equation:
d₂ = (1/2)at²
Substituting the values, we have:
d₂ = (1/2) * (-10 m/s^2) * (0.5 s)²
d₂ = -1.25 m
Note that the negative sign indicates deceleration.
Now, to get the total distance traveled before coming to a stop, we add the distances from both parts:
Total distance = d₁ + d₂
Total distance = 10 m + (-1.25 m)
Total distance = 8.75 m
Therefore, the car will travel 8.75 meters before coming to a stop.