During an auto race, a car with a speed of 85m/s acceleration passed another car rate of 4.0m/s^2 for 3.5s. How far does the car travel during this time?
Okay so I used D= Vi-t+(1/2)t^2
D= 85m-4.0m/s^2+(1/2)3.5^2
D= 87.125 ?
d = Vo*t + 0.5a*t^2.
d = 85*3.5 + 0.5*4*3.5^2 = 322 m.
To find the distance traveled by the car during the given time, we can use the equation of motion:
Distance (D) = Initial velocity (Vi) * time (t) + (1/2) * acceleration (a) * t^2
Given:
Initial velocity (Vi) = 85 m/s
Acceleration (a) = 4.0 m/s^2
Time (t) = 3.5 s
Plugging in these values into the equation, we get:
D = 85 m/s * 3.5 s + (1/2) * 4.0 m/s^2 * (3.5 s)^2
D = 297.5 m + (1/2) * 4.0 m/s^2 * 12.25 s^2
D = 297.5 m + 2.0 m/s^2 * 12.25 s^2
D = 297.5 m + 24.5 m
D = 322 m
Therefore, the car travels a distance of 322 meters during the given time.