A car traveling along a straight road at 66 ft / sec accelerated to a speed of 88 ft / sec over

A car traveling along a straight road at 66 ft / sec...

A car traveling along a straight road at $66 \mathrm{ft} / \mathrm{sec}$ 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?

Key Concepts

Uniformly Accelerated Motion

This concept deals with motion in which acceleration remains constant throughout the duration of the motion. It allows the use of specific kinematic equations to relate displacement, initial velocity, final velocity, acceleration, and time in problems involving objects moving along a straight line.

Kinematic Equations

These are mathematical formulas that describe the relationships among displacement, velocity, acceleration, and time for systems experiencing constant acceleration. One commonly used equation is v² = v?² + 2a?x, which enables the calculation of acceleration when the initial and final velocities and the displacement are known.

Acceleration

Acceleration is a measure of how rapidly an object's velocity changes with time. In the context of constant acceleration, it can be determined by analyzing changes in velocity over displacement or time, making it a fundamental parameter in the study of motion.

Transcript

00:01 Okay, so a car was moving at 66 feet per second, and we increased the speed to 88 feet per second over a distance of 440 feet.

00:18 Okay, and we're asking, we're saying constant acceleration.

00:23 So acceleration equals some constant.

00:29 Of velocity would then be an integral of that, right? because acceleration is the derivative of velocity.

00:41 So velocity is linear with respect to time.

00:48 And then displacement, velocity is the derivative of displacement, is then c over 2 t squared plus some other constant d.

01:03 Okay.

01:04 So what do we know? we know that v of 0 is 66.

01:11 So, oh, sorry, this should be plus d.

01:17 And then, okay, let's not take, let's not find displacement yet.

01:22 V of 0 is 66.

01:24 So that means d is 66.

01:28 So, okay, v of t is ct plus 66.

01:36 Okay, now d of t is c.

01:40 Over to t squared plus 66 t plus some new constant e okay and now we know that d of zero we're just going to call that zero so that means e equals zero if we put in zero over here and um okay and there's some time t we'll call it uh t equals a here then we know that v of a is 88 and d of a and d of a is 440.

02:18 Okay, v of a is 88 gives us that ca plus 66 is 88.

02:24 And d of a is 440 gives us that c over 2a squared plus 66a and e is zero is 440.

02:38 Okay, so now we have two equations and two variables.

02:43 So this should be enough to solve for both big c and a...

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