Question

An object is moving with an initial velocity of 23 m/s. It is then subject to a constant acceleration of 3.5 m/s² for 12 s. How far will it have traveled during the time of its acceleration?

A) 366 m
B) 421 m
C) 474 m
D) 528 m

Answer 1

To find the distance traveled during the time of acceleration, we can use the equation: distance = initial velocity * time + (1/2) * acceleration * time^2. Plugging in the given values, we find that the object will have traveled a distance of 528 m during the time of its acceleration.

Step-by-step explanation:

To find the distance traveled during the time of acceleration, we can use the equation:

distance = initial velocity * time + (1/2) * acceleration * time^2

In this case, the initial velocity is 23 m/s, the time is 12 s, and the acceleration is 3.5 m/s². Plugging these values into the equation, we get:

distance = 23 m/s * 12 s + (1/2) * 3.5 m/s² * (12 s)^2

Simplifying the equation, we get:

distance = 276 m + (1/2) * 3.5 m/s² * 144 s²

Calculating the value, we get:

distance = 276 m + 252 m

distance = 528 m

Therefore, the object will have traveled a distance of 528 m during the time of its acceleration.