Distance using acceleration and time

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A car accelerates from 15 m/s to 28 m/s over 6.4 seconds, and the acceleration is calculated to be 2 m/s². The main challenge is determining the distance traveled during this time, with various incorrect attempts yielding results of 44, 41, 19, and 83 meters. The correct approach involves using the average speed formula or kinematic equations, specifically d = ((vf + vi)/2) * t or d = ut + 1/2 at². Clarification on the use of these formulas helped identify the correct method for calculating distance. Understanding these concepts is crucial for solving similar physics problems effectively.
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Homework Statement


A car accelerates from 15 m/s to 28 m/s in 6.4 s. How far did it travel in this time?

Homework Equations


Unsure- have tried several, and all results have been wrong
Part 1 was to determine acceleration, which is 2 m/(s^2)

The Attempt at a Solution


I've come up with 44, 41, 19, and 83 meters, all of which have been wrong.
 
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mattykay said:
Part 1 was to determine acceleration, which is 2 m/(s^2)
Seems OK

The Attempt at a Solution


I've come up with 44, 41, 19, and 83 meters, all of which have been wrong.
Show what you've done.

There are several ways to solve this. One way is to use the average speed.
 
Do you know your kinematic equations?

Alternatively, if you plot a v-t graph, do you know how to obtain the distance traveled from that graph?
 
I got it; I was using one formula, when I should have been using d=((vf+vi)/2)*t
 
Using d=ut + 1/2 at2 should have given the same answer.
 
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