A car moves along an x axis through a distance of 900 m, starting at rest (at x=0 ) and cnding

A car moves along an x axis through a distance of 900 m,...

A car moves along an $x$ axis through a distance of $900 \mathrm{~m}$, starting at rest (at $x=0$ ) and cnding at rest (at $x=900 \mathrm{~m}$ ). Through the first $\frac{1}{4}$ of that distance, its acceleration is $+2.25 \mathrm{~m} / \mathrm{s}^{2}$. Through the rest of that distance, its acceleration is $-0.750 \mathrm{~m} / \mathrm{s}^{2}$. What are (a) its travel time through the $900 \mathrm{~m}$ and $(\mathrm{b})$ its maximum speed? (c) Graph position $x,$ velocity $v,$ and acceleration $a$ versus time $t$ for the trip.

Key Concepts

Kinematics

Kinematics is the study of motion without considering the forces that cause it. It deals with variables such as displacement, velocity, acceleration, and time, providing the foundational framework to analyze how objects move under various conditions.

Constant Acceleration Equations

These equations describe motion when acceleration is constant. They include relationships such as v = u + at, s = ut + 1/2at², and v² = u² + 2as, which are essential for calculating displacement, final velocity, and time in problems involving uniform acceleration.

Piecewise Motion

Piecewise motion refers to scenarios where an object's journey is divided into segments, each governed by different constant accelerations. This requires analyzing each segment separately using the appropriate kinematic equations and then linking the segments through continuity conditions like matching velocities at their boundaries.

Maximum Speed Determination

Determining the maximum speed in a motion problem, particularly when acceleration changes, involves identifying the point where the object transitions from acceleration to deceleration. At this juncture, the speed is at its peak, and kinematic formulas can be used to quantify that maximum velocity.

Graphical Interpretation of Motion

Graphing motion involves plotting position, velocity, and acceleration as functions of time. These graphs provide visual insights into the nature of the motion, illustrating trends like uniform acceleration or deceleration, and highlighting key points such as maximum speed and changes in the direction of acceleration.

Transcript

00:01 In this problem of motion in a straight line we have given that a car moves along an x -axis through a distance of 900 meters starting from rest that is at initial position this was at x is equal to 0 and ending at rest again so at 900 meter so this was again at rest through the first 4, or we can say 1 4th of the distance, the acceleration is 2 .25 meters per second square.

00:39 So now let's suppose the car moves from starting point which is a and it ends a point which is b.

00:48 Now this length is 900 meters so which is 900 meters.

00:56 Now it is said that through 1 4th of the distance.

01:01 Say this is ab and we can name it as c the acceleration is plus 2 .25 meters per second square and through the rest of the distance the acceleration is minus 0 .75 0 meters per second square now we have to find what are its travel time through 900 meters and also we have to find the maximum speed so let's suppose that the distance from a to c is this is we can name it as delta x1 and this would be 1 fourth so this is 900 meters divide with 4 and the rest of the distance say this is delta x2 and this would be equals to 3 4ths of 900 meters so this is 900 meters now we have to find its travel time through 900 meters so let's suppose that time taken to travel from to c this is t1 and time taken to travel from c to b this is t2 now we have to use the formula that is delta x1 is equal to initial velocity this is v0 t1 plus half a1 and t1 square now putting all the values say delta x1 is equals to 900 meters divided with 4 and initially velocity is equal to 0 so this term is equal to 0 and half multiplied with acceleration this is 2 .25 meters per second square and multiplied with t1 square from here we say that t1 is equals to this is 900 multiplied with 2 and divide with 4 so this is equals to 900 and we have to take the under root also so this is 30 divide with 2 and under root of this is 2 divided with 2 .25 and this would be in seconds.

03:14 So t1 is equal to 14 .14 this is seconds.

03:25 Now we have to calculate t 2.

03:27 So this is delta x2 is equal to...

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