Question
A car traveling along a straight road at $66 \mathrm{ft} /$ sec accelerated to a speed of $88 \mathrm{ft} / \mathrm{sec}$ over a distance of $440 \mathrm{ft}$. What was the acceleration of the car, assuming that it was constant?
Step 1
This can be written as: \[a = \frac{\Delta v}{\Delta t}\] Show more…
Show all steps
Your feedback will help us improve your experience
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
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?
Fundamentals of Algebra
Solving Equations
What constant deceleration would a car moving along a straight road have to be subjected to if it were brought to rest from a speed of $88 \mathrm{ft} / \mathrm{sec}$ in 9 sec? What would be the stopping distance?
Integration
Antiderivatives and the Rules of Integration
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD