A car uniformly accelerates from rest at 3 m/s^2 for 8 seconds. At the end of the 8 seconds the car uniformly brakes at -5m/s^2 for 4 seconds. What total distance is traveled by the car during the 12 seconds of motion?
first 8 seconds:
v = 0 + 3 *8 = 24 m/s at t = 8m/s
x = 0 + 0 t + (1/2)(3)(64) = 96m
the brakes for four seconds
Vi = 24
a = -5
Xi = 96
t = 4
x = 96 + 24(4) - (1/2)(5)(16)
x = 96 + 96 - 40
x = 152
I think this is a two part question, but I don't understand what I am supposed to be solving for:( Do I frist solve for the distance from 0 - 8 seconds and then 9-12 seconds?
Oh....I get it now! Thanks for your help! This has helped me understand many topics in Physics that we have covered this year! Thank you so much! You have no idea how much relief your help has brought me!
You are welcome :)
To find the total distance traveled by the car during the 12 seconds of motion, we need to calculate the distance traveled during the acceleration phase and the distance traveled during the deceleration phase separately. Then, we can add these two distances together to get the total distance traveled.
First, let's find the distance traveled during the acceleration phase.
We know that acceleration, a, is the rate of change of velocity, and the formula is:
a = (change in velocity) / (time)
Given that the car accelerates uniformly from rest, the initial velocity (u) is 0, and the time (t) is 8 seconds. The acceleration (a) is 3 m/s^2.
Using the formula, we can find the change in velocity:
(change in velocity) = a * t = 3 m/s^2 * 8 s = 24 m/s
Next, we can use the formula for distance traveled during uniform acceleration:
distance = (initial velocity * time) + (1/2 * acceleration * time^2)
Since the initial velocity (u) is 0, we can simplify the formula to:
distance = 1/2 * acceleration * time^2
Plugging in the values, we get:
distance = 1/2 * 3 m/s^2 * (8 s)^2 = 1/2 * 3 m/s^2 * 64 s^2 = 96 m
So, during the acceleration phase, the car travels a distance of 96 meters.
Next, let's find the distance traveled during the deceleration phase.
Using the same formula, but now with a negative acceleration of -5 m/s^2 and a time of 4 seconds, we get:
distance = 1/2 * (-5 m/s^2) * (4 s)^2 = 1/2 * (-5 m/s^2) * 16 s^2 = -40 m
Since the distance is a scalar quantity, we take the magnitude of -40 m, which is 40 m.
So, during the deceleration phase, the car travels a distance of 40 meters.
Finally, we can find the total distance traveled by adding the distances during the acceleration and deceleration phases:
total distance = distance during acceleration + distance during deceleration
total distance = 96 m + 40 m = 136 m
Therefore, the total distance traveled by the car during the 12 seconds of motion is 136 meters.