A car accelerates at a rate of 5m/s2 for a period of 4.2 seconds and then moves at a constant speed for 8 seconds more. How far did it travel?
t = 4.2
T = 8
a = 5
I guess it started from a dead stop.
x = (1/2) a t^2 during acceleration
Vi = speed at end of acceleration = a t
X = Vi T during constant speed so
distance = (1/2) a t^2 + Vi T
212
To find the total distance traveled by the car, we need to calculate the distance covered during acceleration and the distance covered at constant speed, and then add them together.
1. Distance during acceleration:
During acceleration, the car's initial velocity is 0 m/s (assuming it starts from rest). The formula for distance traveled during acceleration is given by: distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Given:
- Initial velocity (u) = 0 m/s
- Time taken (t) = 4.2 seconds
- Acceleration (a) = 5 m/s^2
Plugging these values into the formula, we have:
distance = (0 * 4.2) + (0.5 * 5 * 4.2^2)
distance = 0 + (0.5 * 5 * 17.64)
distance = 0 + (0.5 * 88.2)
distance = 0 + 44.1
distance = 44.1 meters
2. Distance at constant speed:
The car moves at a constant speed for 8 seconds, so the distance covered is given by: distance = speed * time
Given:
- Speed (v) = Constant speed
- Time taken (t) = 8 seconds
We don't know the constant speed, but we don't need it to find the total distance traveled.
distance = speed * time
Now, adding both distances together:
total distance = distance during acceleration + distance at constant speed
total distance = 44.1 meters + distance at constant speed
We don't have the value of the distance at constant speed, so we cannot calculate the total distance without that information. Please provide the speed at which the car moves at a constant speed.