Distance covered by car while decelerating

[11NEPHILEM11]

Homework Statement

A car level on ground, starting from 15m/s, decelerates uniformly to rest in 20 seconds. What distance does the car cover while decelerating?

Homework Equations

Vf = Vi+at
Vf^2=Vi^2+2aX
X=Vit+(1/2)at^2

X = Distance
Vf = Final velocity
Vi = Initial velocity
a = acceleration
t = time

The Attempt at a Solution

I started with the X=Vit+(1/2)at^2 formula and attempted to plug in the numbers but I have no idea how to find acceleration. I mean, I know acceleration is Δv over Δt but how am I supposed to find the final velocity when I need the acceleration to plug in the formula? I'll admit I zoned out a little in my last class because I was tired from work the day before.

 

[Simon Bridge]

Since you only have three kinematic equations, you will need to use two of them.
Note - you have initial and final velocities, and the time ... what is the acceleration?

The other approach is to sketch a velocity-time graph.
The distance covered is the area under the graph ... which will be a triangle.
You know how to find the area of a triangle right?

 

[SteamKing]

Are you sure you haven't listed an equation above, where you know 1. the initial velocity, 2. the final velocity, 3. the duration for the change in velocities, and 4. which leaves only acceleration (or deceleration) to be determined?

You have to understand what your relevant equations mean.

 

[11NEPHILEM11]

If I am understanding correctly, I already have my initial and final velocity? If my initial velocity is 15m/s then what is my final velocity? 0?

 

[Simon Bridge]

If I am understanding correctly, I already have my initial and final velocity? If my initial velocity is 15m/s then what is my final velocity? 0?

Easy to check - the question says that the car "decelerates to rest" ... so what does that tell you about the final velocity?