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(II) The position of an object is given by $x=A t+B t^{2}$ ,where $x$ is in meters and $t$ is in seconds. $(a)$ What are theunits of $A$ and $B ?(b)$ What is the acceleration as a functionof time? (c) What are the velocity and acceleration at$t=5.0 \mathrm{s} ?$ (d) What is the velocity as a function of time if$x=A t+B t^{-3} ?$

a) $m / s$ $m / s^2$b) 2$B m / s^2$c) $(A+10 B) m / s$ 2$B m / s^2$d) $v=A-3 B t^{-4}$

Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

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so here for part A. Since the units of a times, the units of tea must equal meters. So we have a times t Uh this means that eight times team must equal a unit of meters. So this means that we know that t is measuring seconds. Therefore, a must be meters per second multiplied by seconds in order to equal meters there, four units of a should equal meters per second for be. We know that beat must equal meters, which means that this is gonna be meters per second squared multiplied by second squared, which means that the units of B must equal meters per second squared for part B. They want us to find the acceleration. So the acceleration is going to be the second derivative off the position function. We Excuse me, we know that for part B, the position function equals 80 plus B t squared. And so the velocity would be equal to the first derivative. This would be a plus two b t and then for the acceleration. This is gonna equal the derivative of the velocity function. With respect to time, this would be equal to the second derivative of the position function again with respect to time and this would be equal to simply two times. Be, uh, the units here would be meters per second squared. Now for part C, get a new workbook for part C. They want us to find the velocity at the velocity at five seconds. So we can say that V equals a plus two times B T, which means that velocity at five seconds would be equal to a plus two times five times be so this would be equal to a plus 10 B units here would be meters per second and then the acceleration, um, would be constant at to be and the units here would be meters per second squared. And finally, for part D, they want us to find the velocity function. We know the position function is a T plus B t to the negative third power. Therefore, the velocity function will be the first derivative of the position function, so this would be equal to a minus three b times t to the negative fourth power. That is the end of the solution. Thank you for watching

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