MHB How Does Velocity Impact Acceleration in Race Cars?

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The discussion focuses on calculating the acceleration of a race car given its velocity function, which is defined as v_x(t) = (0.930 m/s^3)t^2. The acceleration is derived using the formula A = dv/dt, resulting in A = 1.86t. When the velocity reaches 14.5 m/s, the time calculated is approximately 3.94 seconds, leading to an acceleration of about 7.35 m/s². There is a minor discrepancy noted in the velocity function, questioning whether it should be 0.930 or 0.933. The calculations emphasize the relationship between velocity and acceleration in the context of race car dynamics.
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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

$A(3.94)=1.86\cdot3.94=7.32 m/s^2$
 
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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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