Answer :
The speed at the end of 5 seconds is 40.00 m/s, the average speed during this period is 20.00 m/s, and the distance traveled is 100.00 m, illustrating basic principles of motion under constant acceleration.
The question involves an object that starts from rest and undergoes constant acceleration along a straight line. Specifically, it asks for the speed at the end of 5 seconds, the average speed during this interval, and the distance traveled in the same period given a constant acceleration of 8.00 m/s2.
- To find the speed at the end of 5 seconds, you can use the formula v = u + at, where v is the final velocity, u is the initial velocity (0 m/s since it starts from rest), a is the acceleration, and t is the time. Substituting the given values, we get v = 0 + (8.00 m/s2)(5.00 s) = 40.00 m/s.
- For the average speed, we take the formula average speed = total distance / total time. However, in cases of constant acceleration from rest, the average speed is simply half of the final speed, so average speed = 20.00 m/s.
- To calculate the distance traveled, use the formula s = ut + (1/2)at2, replacing the values gives us s = (1/2)(8.00 m/s2)(5.00 s)2 = 100.00 m.
Through this calculation, we find the speed after 5 seconds to be 40.00 m/s, the average speed across this time interval to be 20.00 m/s, and the distance covered to be 100.00 m, demonstrating the concepts of acceleration and motion in physics.
