A car traveling at 14 m/s accelerates at 2.11 m/s2 for 14 seconds. To the nearest meter how far does it travel? To the nearest tenth of a m/s, what is the final velocity of the car in the previous problem?
To find the distance traveled by the car, we can use the kinematic equation:
distance = initial velocity * time + (0.5 * acceleration * time^2)
Given:
Initial velocity (u) = 14 m/s
Acceleration (a) = 2.11 m/s^2
Time (t) = 14 s
Substituting these values into the equation:
distance = (14 * 14) + (0.5 * 2.11 * 14^2)
distance = 196 + (0.5 * 2.11 * 196)
distance = 196 + (207.64)
distance = 403.64
Therefore, the car travels approximately 404 meters.
To find the final velocity of the car, we can use the equation:
final velocity (v) = initial velocity + (acceleration * time)
Given:
Initial velocity (u) = 14 m/s
Acceleration (a) = 2.11 m/s^2
Time (t) = 14 s
Substituting these values into the equation:
final velocity = 14 + (2.11 * 14)
final velocity = 14 + 29.54
final velocity = 43.54
Therefore, the final velocity of the car is approximately 43.5 m/s.