NEXT STANDARD VEW PRINTER VERSION BACK SIGNMENT RESOURCES omework 2 Chapter 02,

Question

NEXT STANDARD VEW PRINTER VERSION BACK SIGNMENT RESOURCES omework 2 Chapter 02, Problem 015 Your answer is partially correct. Try again 209 029 002 A particle's position is given by x 14.0-12.00t32, in which x is in meters and t is in seconds. (a) What is its velocity at t- 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) ever an instant when the velocity is zero? If so, give the time t; if not, answer "o. (r) Is there a time after t-3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer "O" (a) N (b)T negative (e) N (d)T decreasing (e) Number2 UnitsT m/s 030 042 m/s s Interactive Exercises 2.03 Average Velosisy Interactive Exercises Interactive Exercises 2.111 Free FalL(NASA SHOW HINT Review Score Review Results by Study objective LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM MATH HELP LINK TO TEXT

Explanation / Answer

I'm assuming you only need answer to part C. If you need help with any other parts comment below.

x = 14 - 12t + 3t^2

Velocity is given by

V = dx/dt

V = d[14 - 12t + 3t^2]/dt

V = 0 - 12*1 + 3*2*t

V = 6t - 12

Now at t = 1 sec Velocity will be

V = 6*1 - 12 = -6 m/sec

Now Remember that Velocity is a vector quantity, but speed is a scalar quantity, SO speed will not have any direction, it will only have magnitude

Speed = |Velocity|

Speed = |-6|

Speed = 6 m/sec

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