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3 years ago

Gragh h(t) = 400 − 39t − 157e−t/4

Mathematics

1 answer:

Paraphin [41]3 years ago

3 0

The graph of $h(t) = 400 - 39t - 157e^{ \frac{t}{4} }$ is attached.

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Analysis of an accident scene indicates a car was traveling at a velocity of 69.5 mph (31.1 m/s) along the positive x-axis at th

STatiana [176]

We can find the acceleration via

${v_f}^2-{v_i}^2=2a\Delta x$

We have

$\left(5.20\dfrac{\rm m}{\rm s}\right)^2-\left(31.1\dfrac{\rm m}{\rm s}\right)^2=2a(115\,\mathrm m)$

$\implies\boxed{a=-4.09\dfrac{\rm m}{\mathrm s^2}}$

Then by definition of average acceleration,

$a_{\rm ave}=\dfrac{v_f-v_i}t$

so that

$-4.09\dfrac{\rm m}{\mathrm s^2}=\dfrac{5.20\frac{\rm m}{\rm s}-31.1\frac{\rm m}{\rm s}}t$

$\implies\boxed{t=6.33\,\mathrm s}$

We alternatively could have found the time without knowing the acceleration. Since acceleration is constant, the average velocity is

$v_{\rm ave}=\dfrac{x_f-x_i}t=\dfrac{v_f+v_i}2$

Then

$\dfrac{115\,\rm m}t=\dfrac{5.20\frac{\rm m}{\rm s}+31.1\frac{\rm m}{\rm s}}2$

$\implies\boxed{t=6.33\,\mathrm s}$

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