A car traveling at 18 m/s accelerates at 5.5 m/s2 for 9 s. How far does the car travel during the time it is accelerating for 9 seconds?
1. Change in velocity = acceleration * time = 5.5 m/s2 * 9 s = 49.5 m/s
2. Final velocity = Initial Velocity + Change in velocity = 18 + 49.5 m/s = 67.5 m/s
3. distance = (Final velocity ^ 2 - Inital velocity ^ 2) / (2 * acceleration) = (67.5 m/s * 67.5 m/s - 18 * 18) / (2 * 5.5 m/s2) = 384.75 m
To find the distance traveled during the time the car is accelerating, we can use the equation of motion:
distance = initial velocity * time + 0.5 * acceleration * time^2
Given:
Initial velocity (u) = 18 m/s
Acceleration (a) = 5.5 m/s^2
Time (t) = 9 s
Substituting these values into the equation:
distance = 18 m/s * 9 s + 0.5 * 5.5 m/s^2 * (9 s)^2
Let's calculate it step by step:
1. Multiply the initial velocity by the time:
18 m/s * 9 s = 162 m
2. Calculate the squared value of the time:
(9 s)^2 = 81 s^2
3. Multiply the acceleration by half of the squared time:
0.5 * 5.5 m/s^2 * 81 s^2 = 221.25 m
4. Add the results from step 1 and step 3:
162 m + 221.25 m = 383.25 m
Therefore, the car travels approximately 383.25 meters during the time it is accelerating for 9 seconds.