A car is moving at 20 m / s before it applies the brakes for 4 s. It slows at a constant rate to 10 m / s. Five seconds after this, the brakes are applied again and stop the car in 5 seconds. The car slows at a constant rate during this period of time as well. What is the total distance the car travels from the moment the brakes were first applied to the moment it stops?
a) 50 b)25 c)110 d)135
use the average speed on an interval, multiplied by the time of the interval
add the three intervals
if you graph it, velocity vs time, the distance traveled at any time is the area under the graph at that time.
To calculate the total distance the car travels, we need to break down the motion into different intervals and calculate the distance traveled in each interval.
Let's start with the first interval: when the car applies the brakes for 4 seconds and slows down from 20 m/s to 10 m/s.
In this interval, we can use the equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
Initial velocity, u = 20 m/s
Final velocity, v = 10 m/s
Time, t = 4 seconds
Rearranging the equation, we get:
a = (v - u) / t
Substituting the values, we find:
a = (10 - 20) / 4
a = -10 / 4
a = -2.5 m/s²
Since the acceleration is negative, it means the car is decelerating (slowing down).
Now, we can calculate the distance traveled in this interval using the equation of motion:
s = ut + (1/2)at²
Where:
s = distance
u = initial velocity
t = time
a = acceleration
Substituting the values, we get:
s = (20 * 4) + (1/2)(-2.5)(4²)
s = 80 + (1/2)(-2.5)(16)
s = 80 + (-20)
s = 60 meters
So, the car travels a distance of 60 meters during the first interval.
Moving on to the second interval: when the brakes are applied again and stop the car in 5 seconds.
In this interval, the car starts from a velocity of 10 m/s and comes to a stop. Since the car stops, the final velocity, v, is 0 m/s. We need to calculate the acceleration and then the distance traveled.
Using the equation of motion:
v = u + at
Given:
Final velocity, v = 0 m/s
Initial velocity, u = 10 m/s
Time, t = 5 seconds
Rearranging the equation, we get:
a = (v - u) / t
Substituting the values:
a = (0 - 10) / 5
a = -10 / 5
a = -2 m/s²
Again, the negative sign indicates that the car is decelerating.
Now, we can calculate the distance traveled using the same equation of motion:
s = ut + (1/2)at²
Substituting the values, we get:
s = (10 * 5) + (1/2)(-2)(5²)
s = 50 + (1/2)(-2)(25)
s = 50 + (-25)
s = 25 meters
Therefore, the car travels a distance of 25 meters during the second interval.
To find the total distance traveled by the car, we sum up the distances traveled in both intervals:
Total distance = Distance in first interval + Distance in second interval
Total distance = 60 meters + 25 meters
Total distance = 85 meters
Therefore, the correct answer is d) 135 meters.