A proton moves along the x axis according to the equation...

A proton moves along the $x$ axis according to the equation $x=50 t+10 r^{2},$ where $x$ is in meters and $t$ is in seconds. Calculate (a) the average velocity of the proton during the first $3.0 \mathrm{~s}$ of its motion. (b) the instantancous velocity of the proton at $t=3.0 \mathrm{~s}$, and (c) the instantancous acceleration of the proton at $t=3.0 \mathrm{~s}$. (d) Graph $x$ versus $t$ and indicate how the answer to (a) can be obtained from the plot. (c) Indicate the answer to (b) on the graph. (f) Plot $v$ versus $t$ and indicate on it the answer to (c).

A proton moves along the $x$ axis according to the equation $x=50 t+10 r^{2},$ where $x$ is in meters and $t$ is in seconds. Calculate (a) the average velocity of the proton during the first $3.0 \mathrm{~s}$ of its motion.
(b) the instantancous velocity of the proton at $t=3.0 \mathrm{~s}$, and (c) the instantancous acceleration of the proton at $t=3.0 \mathrm{~s}$. (d) Graph $x$ versus $t$ and indicate how the answer to (a) can be obtained from the plot. (c) Indicate the answer to (b) on the graph. (f) Plot $v$ versus $t$ and indicate on it the answer to (c).

Key Concepts

Differentiation in Kinematics

Differentiation is a mathematical tool used in kinematics to relate position, velocity, and acceleration. By differentiating the position function with respect to time, one obtains the velocity function, and further differentiation of the velocity function yields the acceleration. This process is critical for analyzing and understanding the changing dynamics of moving objects.

Graphical Analysis of Motion

Graphical analysis in kinematics involves plotting the position, velocity, or acceleration as functions of time to visualize motion. The position versus time graph can be used to determine average velocity by calculating the slope between two points, while the tangent at any point gives the instantaneous velocity. Similarly, a velocity versus time graph can display acceleration as the slope at a specific time.

Instantaneous Velocity

Instantaneous velocity is the velocity of an object at a specific moment in time. It is obtained by taking the derivative of the position function with respect to time, which provides the slope of the tangent line to the position versus time graph at any given point.

Average Velocity

Average velocity is defined as the displacement (change in position) divided by the elapsed time over a time interval. It gives an overall measure of the rate at which an object changes its position between two moments and indicates the net effect of varied speeds or directions during the interval.

Instantaneous Acceleration

Instantaneous acceleration is the rate of change of instantaneous velocity with respect to time. It is calculated by differentiating the velocity function with respect to time, yielding the slope of the velocity versus time graph at a particular moment.

Transcript

00:01 In this problem of motion along a straight line we have given that a proton moves along the x -axis according to the equation which is x is equals to 50 multiplied with t plus 10 t square where x is in meters and t is in seconds.

00:15 We have to calculate the average velocity of the proton during first three seconds of its motion.

00:21 So average velocity that means the we can say we average so position change so x at 3 minus x at 0 divided with the time taken which is equal to delta t is equal to three second now putting x is equal to 3 so 50 multiplied with 3 which is 150 and 10 multiplied with 3 so this is equals to 50 multiplied with 3 plus 10 multiplied with 3 square which is equal to 9 t minus x 0 putting 0 so this is equal to 0 and time t is equal to 3 seconds so when we solve it this is equals to 80 meters per second this is the answer for part a now part b says the instantaneous velocity of the proton at t is equal to three seconds so velocity as a function of t is equal to when we differentiate it so 50 minus 20 t now we have to find the instantaneous velocity so we have to put v is equal to here t is equal to three second now putting the value so 50 minus 20 multiplied with three so this is equals to and here this value would be plus because this is plus so when we add it so this is 50 plus 60 which is equal to 110 meters per second this is the answer for part b now part c in part c we have to find the instantaneous acceleration of the proton at t is equal to 3 second now we have to find the acceleration in terms of t so this is equal to differentiation of this term so this is simply 20 so acceleration is constant so we say that this is only 20 meters per second is square so acceleration is equal to 20 meters per second square so we say that answer for part a for part b and answer for part c now part d.

02:12 Now part d says we have to graph x versus t and we have to indicate it on graph so here is the graph i'm showing you...

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