Question

The average acceleration of an object is given by a = where a is the average acceleration, v is the velocity, and t is the time. Find the velocity after 6.8 s of an object whose acceleration is 18 ft/s^2 (feet per second squared).

a) 18 ft/s
b) 122.4 ft/s
c) 122.4 ft
d) 108 ft/s

Answer 1

Using the formula v = u + at, and given that the initial velocity u is zero, the object's velocity after 6.8 seconds with a constant acceleration of 18 ft/s^2 is calculated to be 122.4 ft/s.

Step-by-step explanation:

The average acceleration of an object a is the change in velocity v over the time interval t. For an object with a constant acceleration, the final velocity v after a time t can be calculated using the formula v = u + at, where u is the initial velocity, which is zero in this case because it is not provided, making the initial velocity insignificant for this specific calculation.

Given an acceleration of 18 ft/s2 over a period of 6.8 seconds, one can calculate the velocity at the end of this time interval using the following steps:

1. Calculate the change in velocity: Δv = a × t
2. Δv = 18 ft/s2 × 6.8 s
3. Δv = 122.4 ft/s

This means that after 6.8 seconds, the object's velocity is 122.4 ft/s.