Acceleration of Sports Car: Solve Problem

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Homework Help Overview

The problem involves a sports car accelerating from rest to a speed of 100 km/hr within a time frame of 6.2 seconds, with the goal of determining its acceleration in meters per second squared (m/s²).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss converting units from km/hr to m/s and explore the appropriate equations for calculating acceleration. There are attempts to clarify the relationship between velocity, time, and acceleration, with some participants questioning the accuracy of initial calculations.

Discussion Status

The discussion includes various approaches to the problem, with some participants providing guidance on relevant equations and unit conversions. There is an acknowledgment of differing interpretations of the calculations, and while some participants express uncertainty, others seem to gain clarity from the exchanges.

Contextual Notes

One participant notes the challenge of not having a textbook for reference, which may impact their understanding of the concepts being discussed.

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Homework Statement



A sports car accelerates from rest to 100km/hr in 6.2s. What is it's acceleration in m/s^2?

Homework Equations


The Attempt at a Solution


I converted the 100 km/hr to m/s and got 16.7 m/s. I think that the formula that I am supposed to use is Vf-Vo/t. I don't know what to do next or if even what I did is correct.
 
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100 km/hr=(100x1000)meter/60min=(100x1000)meter/(60x60)sec=27.8meter/sec.

Now look at the units for acceleration. This should tell you what you must do with velocity(m/s) to get m/s^2.
 
100 kM/hr = 27.78 m/s not 16.7 m/s. Look in your book. You should find a reference equation that looks something like v=at+vo. Since you know v0=o at time=0 the equation becomes simple. Now solve for acceleration since you have been given both the time and final velocity.

Hope this helps.
 
since 100km/h = 27.8 m/s

use v^2 = V0^2 + 2a(y-y0)

27.8^2 = 0 + 2a(y) ; 2 unknowns
/////(27.8^2) / 2a = y

then use y = y0+v0T+.5aT^2
plug in with the Y you solved for in the other equation.
772.84 / 2a = 0 + 0T + .5aT^2
remember t = 6.2
solve for a; a = 4.48 m/s^2

check my math, i don't know
 
Last edited:
rsala said:
since 100km/h = 27.8 m/s

use v^2 = V0^2 + 2a(y-y0)

27.8^2 = 0 + 2a(y) ; 2 unknowns
/////(27.8^2) / 2a = y

then use y = y0+v0T+.5aT^2
plug in with the Y you solved for in the other equation.
772.84 / 2a = 0 + 0T + .5aT^2
t = 6.2
solve for a; a = 4.48 m/s^2

check my math, i don't know
You are making it too complicated. You have one equation with one unknown: v=at+vo
 
I'm not sure if I exactly understand. I understand a little more than I did before. It is hard for me because my class does not use a book and we have one practice problem as our notes.
 
Last edited:
This (Vf-Vo/t) is what you showed in your first post for finding the acceleration a. And as Newton1Law wrote, "Since you know v0=o at time=0..." then a=Vf/t. That's all there is to it.
 
Thank you. That makes sense to me now.
 

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