In an arcade video game, a spot is programmed to move across...

In an arcade video game, a spot is programmed to move across the screen according to $x=9.00 t-0.750 t^{3}$, where $x$ is distance in centimeters measured from the left edge of the screen and $t$ is time in seconds. When the spot reaches a screen edge, at either $x=0$ or $x=15.0 \mathrm{~cm}, t$ is reset to 0 and the spot starts moving again according to $x(t)$. (a) At what time after starting is the spot instantaneously at rest? (b) $A t$ what value of $x$ does this occur? (c) What is the spot's acceleration (including sign) when this occurs? (d) 1st it moving right or left just prior to coming to rest? (e) Just after? (f) At what time $t>0$ docs it first reach an edge of the screen?

In an arcade video game, a spot is programmed to move across the screen according to $x=9.00 t-0.750 t^{3}$, where $x$ is distance in centimeters measured from the left edge of the screen and $t$ is time in seconds. When the spot reaches a screen edge, at either $x=0$ or $x=15.0 \mathrm{~cm}, t$ is reset to 0 and the spot starts moving again according to $x(t)$. (a) At what time after starting is the spot instantaneously at rest? (b) $A t$ what value of $x$ does this occur? (c) What is the spot's acceleration (including sign) when this occurs? (d) 1st it moving right or left just prior to coming to rest? (e) Just after?
(f) At what time $t>0$ docs it first reach an edge of the screen?

Key Concepts

Boundary Conditions in Kinematics

Boundary conditions specify the limits within which an object's motion occurs, such as fixed edges or endpoints. In problems where an object moves within constrained limits, these conditions are used to determine when the object reaches a boundary, which may trigger additional events such as resetting the motion or changing the dynamic behavior.

Critical Points in Kinematics

Critical points occur when the derivative of the position function (i.e., the velocity) is zero. These points are significant because they often correspond to changes in the direction of motion or moments when the object is instantaneously at rest. Identifying and analyzing these points is key in studying the dynamics of motion.

Direction Change Analysis

The analysis of the sign of velocity before and after a moment of rest is used to determine whether the object changes direction. By comparing the velocity values just prior to and immediately after the moment when it is zero, one can infer the change in direction. This analysis is crucial in understanding the complete behavior of moving objects.

Velocity

Instantaneous velocity is the time derivative of the position function. It gives both the speed and the direction of an object at a specific moment in time. When the velocity is zero, the object is momentarily at rest. Understanding velocity is essential for determining moments of rest or changes in direction during motion.

Acceleration

Acceleration is the derivative of the velocity with respect to time or equivalently the second derivative of the position function. It describes how the velocity of an object changes over time. Analyzing acceleration, including its sign, helps determine whether an object is speeding up or slowing down at any given moment.

Position Function

A position function represents the location of an object as a function of time. In kinematics, it is used to describe how an object's displacement changes over time, and by knowing this function, one can derive information about the object's motion, such as its speed, direction, and acceleration.

Transcript

00:01 In this problem of motion along a street line we have given that in an arcade video game i spot is programmed to move across the screen according to x is equal to 9 .0 this is 9 .0 t t minus 0 .750 t cube 0 .750 t cube and where x is distance in centimeter so x is in centimeter so x is in centimeters measured from the left edge of the screen and time t is in seconds so time t is in seconds when the spot reaches screen edge at either x is equal to 0 or x is equal to 15 centimeters either x is equal to 0 or x is equal to 15 .0 centimeters t is reset to 0 so again time would be reset at t is equal to to 0 and the spot starts moving again according to the formula that is x t is equals to 9 .0 t minus 0 .750 tq now we have to find for part a at what time after starting is spot instantaneously at rest so for this we have to find the velocity so velocity say vt is equals to this is 9 minus this value is after differentiation so this value would be 9 minus 9 divide with 4 t square now this is the velocity in terms of t the particle would be just when this is equal to 0 now putting the value say t square would be equal to sending to the left -hand side so t is equal to 4 that means t is equal to plus and minus 2 so we have to take time positive because time cannot be negative so t is equal to 2 .0 seconds so this is the solution for part a that is 2 seconds now for part b so here part b part we say at what value of x does this occur so we have to find the value so putting the value t is equal to 2 in this formula so 9 multiplied with 2 which is 18 minus 0 .75 and 2 cube when we solve it this is equal to this is this is equals to 12 centimeter that means 12 .0 centimeter from left edge of a screen so answer is 12 .0 centimeter from left edge of a screen so this is the answer for part b now part c in part c we have to find what is the sports acceleration including sign when this occurs.

03:18 So first we have to find the acceleration.

03:20 So velocity is here.

03:21 So we have to find the acceleration in terms of t.

03:25 D differentiating it so this value would be minus 9 divided with 2.

03:29 Minus 9 divided with 2 t.

03:32 So this is the acceleration in terms of t.

03:35 Now we have to put t is equal to 2 seconds.

03:38 So acceleration is equal to putting t is equal to 2 seconds.

03:44 So this value would be a2.

03:45 So, we have a 2 is equal to 9 this is minus 9 divided with 2 multiplied with 2 which is equals to minus 9 .0 cm per second square so this is the acceleration at this time t is equal to 2 seconds now part d so for part d we have given that is it moving right or left just prior to coming to rest since velocity is equal to at t is equal to two second velocity putting the value so this is equals to 0 and before coming to rest velocity is greater than 0 then the spot had been moving towards right towards so we say that spot is moving right so we say that answer is spot is moving towards right so this is the answer now for part e.

04:59 In part e, this is part e.

05:05 And now we have to find just after...

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