Question
A car traveling along a straight road at $66 \mathrm{ft} / \mathrm{sec}$ accelerated to a speed of $88 \mathrm{ft} / \mathrm{sec}$ over $\mathrm{a}$ distance of $440 \mathrm{ft}$. What was the acceleration of the car,assuming it was constant?
Step 1
First, we need to find the time it took for the car to accelerate from 66 ft/sec to 88 ft/sec. To do this, we can use the formula for acceleration: acceleration = (final velocity - initial velocity) / time. We need to solve for time. Show more…
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The distance $s$ (in feet) covered by a car traveling along a straight road is related to its initial speed $u$ (in $\mathrm{ft} / \mathrm{sec})$, its final speed $v$ (in $\mathrm{ft} / \mathrm{sec})$, and its (constant) acceleration $a$ (in $\mathrm{ft} / \mathrm{sec}^{2}$ ) by the equation $v^{2}=u^{2}+2 a s$. a. Solve the equation for $a$ in terms of the other variables. b. A car starting from rest and accelerating at a constant rate reaches a speed of $88 \mathrm{ft} / \mathrm{sec}$ after traveling $\frac{1}{4}$ mile $(1320 \mathrm{ft})$. What is its acceleration?
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