Question

A dragster can achieve average accelerations of 16.0 m/s^2. Suppose such a dragster accelerates from rest at this rate for 2.56 s. What is the distance traveled and the final velocity of the dragster in this time?

Answer 1

The dragster travels a distance of 52.224 meters and reaches a final velocity of 40.96 m/s after accelerating from rest at 16.0 m/s^2 for 2.56 seconds.

Step-by-step explanation:

The question asks for the distance traveled and the final velocity of a dragster which accelerates from rest with an average acceleration of 16.0 m/s2 for a time period of 2.56 seconds. To find the distance traveled by the dragster, we can use the kinematic equation for uniformly accelerated motion:

Distance (d) = (initial velocity × time) + (½ × acceleration × time2)

In this case, the initial velocity is 0 m/s because the dragster starts from rest, so the equation simplifies to:

d = ½ × (16.0 m/s2) × (2.56 s)2 = 52.224 meters

To find the final velocity, we also use a kinematic equation for uniformly accelerated motion:

Final velocity (vf) = initial velocity + (acceleration × time)

Since the initial velocity is 0 m/s, the final velocity is simply:

vf = 0 m/s + (16.0 m/s2 × 2.56 s) = 40.96 m/s

Therefore, the dragster travels a distance of 52.224 meters and reaches a final velocity of 40.96 m/s after 2.56 seconds of acceleration.