How Does Velocity Impact Acceleration in Race Cars?

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SUMMARY

The discussion focuses on the relationship between velocity and acceleration in race cars, specifically analyzing a car's motion with a velocity function defined as \(v_x(t) = (0.930 \, \text{m/s}^3)t^2\). The acceleration is calculated using the formula \(A = \frac{dv}{dt} = 1.86t\). When the velocity reaches 14.5 m/s, the time is determined to be approximately 3.94 seconds, resulting in an acceleration of 7.32 m/s². A discrepancy in the coefficient of the velocity function (0.930 vs. 0.933) was noted, prompting clarification on the correct value.

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karush
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A race car strarts from rest and travels east along a straight and level track.
For the firs $5.0s$ of the cars motion,
the eastward compent of the car's velocity,is given by
$v_x(t)=(0.930 m/s^3)t^2 $
What is acceleration of the car when $v_x=14.5$

ok my first step with this was

$$A=\frac{dv}{dt}=1.86t$$
if $v=14.5=0.933t^2$
then
$$t=3.94$$

so

$A(3.94)=1.86\cdot3.94=7.32 m/s^2$
 
Last edited:
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You give, as the speed function, [math]0.930 t^2[/math], but when calculating the time when the speed is 14.5, you have [math]0.933t^2[/math]. Which is correct, 0.930 or 0.933?
 
Hope this is better got lost in the W|A output

A race car strats from rest and travels east along a straight and level track.
For the first 5.0s of the cars motion,
the eastward compent of the car's velocity,is given by
$v_x(t)=(0.93 m/s^3)t^2$
a. What is acceleration of the car when $v_x=14.5
$$\begin{align*} \displaystyle
A&=\frac{dv}{dt}=1.86t\\
v&=14.5=0.93t^2\\
t&=3.94\\
A(3.94)&=1.86\cdot3.94=7.35 m/s^2
\end{align*}$$

I was ? on round off
 

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