# Acceleration of Sports Car: Solve Problem

• sprinter08
In summary, the conversation is about finding the acceleration of a sports car that accelerates from rest to 100km/hr in 6.2 seconds. The correct formula to use is v=at+vo and since the initial velocity (vo) is 0, the formula simplifies to a=Vf/t. Therefore, the acceleration of the car is 4.48 m/s^2.
sprinter08

## Homework Statement

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

## 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.

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.

## 1. How does acceleration affect the performance of a sports car?

The acceleration of a sports car is a crucial factor in its overall performance. It determines how quickly the car can reach its maximum speed and how efficiently it can accelerate from a standstill. A higher acceleration means the car can cover more distance in a shorter amount of time, resulting in better performance.

## 2. What factors contribute to the acceleration of a sports car?

The acceleration of a sports car is influenced by several factors, including the engine power, weight of the car, aerodynamics, and traction. A more powerful engine with a better power-to-weight ratio, a lighter weight, streamlined design, and good grip on the road can all contribute to a faster acceleration.

## 3. How can the acceleration of a sports car be measured?

The acceleration of a sports car can be measured using a variety of methods, including using a stopwatch to time how long it takes the car to reach a certain speed, using a dynamometer to measure the power output of the engine, or using a data logger to track the car's acceleration in real-time.

## 4. Are there any safety concerns related to the acceleration of sports cars?

Yes, there are some safety concerns related to the acceleration of sports cars. A higher acceleration can result in a faster speed, which can increase the risk of accidents if the driver is not skilled or responsible. It is important for drivers to follow speed limits and drive responsibly to ensure their safety and the safety of others on the road.

## 5. How can the acceleration of a sports car be improved?

The acceleration of a sports car can be improved by making modifications to the engine, reducing the weight of the car, and improving the aerodynamics. Upgrading the engine components, such as the air intake and exhaust system, can increase the power and torque, resulting in better acceleration. Removing unnecessary weight, such as extra seats or heavy accessories, can also improve the power-to-weight ratio. Additionally, making small changes to the car's design, such as adding spoilers or diffusers, can improve its aerodynamics and reduce drag, resulting in faster acceleration.

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