Question

|| A particle’s position on the x-axis is given by the function x = (t 2 - 4t + 2) m, where t is in s. a. Make a position-versus-time graph for the interval 0 s … t … 5 s. Do this by calculating and plotting x every 0.5 s from 0 s to 5 s, then drawing a smooth curve through the points. b. Determine the particle’s velocity at t = 1.0 s by drawing the tangent line on your graph and measuring its slope. c. Determine the particle’s velocity at t = 1.0 s by evaluating the derivative at that instant. Compare this to your result from part b. d. Are there any turning points in the particle’s motion? If so, at what position or positions? e. Where is the particle when vx = 4.0 m/s?

Answer 1

a )

x = (t ² - 4t + 2) m, where t is in s.

a. Make a position-versus-time graph for the interval 0 s … t … 5 s. Do this by calculating and plotting x every 0.5 s from 0 s to 5 s,

at t = 0s , x = 2 m

t = 0.5 s , x = 0.25m

t = 1 s,x = 1m

t = 1.5s , x = 1.75 m

t = 2s , x = 2 m

t = 2.5s , x = 1.75m

t = 3 s ,x = 1m

t = 3.5 s , 0.25m = x

t = 4 s , x = 2m

t = 4.5 s , x = 4.25 m

t = 5 s , x = 7m

b )

v = dx / dt

v = 2t - 4 , where x = t² - 4 t + 2

t = 1s

v = 2 - 4

v = - 2 m /s

as we can write v = 2 m/ s

c )

v = |dx/dt|_t=1s

v = |2t-4|_{t = 1s}

v = 2m/s

same as part b

d )

there are turing pointas in the motion when ever we submitted t value in part a

we got some minus values are written positive

e )

v_x = 4 m /s

v_x = 2t- 4 = 4

2t = 8

t = 4 sec

at t = 4sec particle's velocity will be 4 m /s