Figure (3-Q1) shows the x- coordinate of a particle as a function of time. Find the signs of vx and ax at t=t1, t= t2 and t=t3.

Question 1

Question 2

If x' and x" be the x-coordinate of the particle at initial time t' and t" respectively then vx =(x"-x')/(t"-t') = tanθ.

For t"-t' infinitesimally small it is the vx at that instant.

So slope of the tangent at any point in the above graph gives vx .

At t=t1, tanθ is positive, so sign of vx is positive.

At t= t2 the slope of the curve is horizontal, so tanθ=0 → vx =0.

At t=t3 the slope of the curve is negative, so sign of vx is negative.

kratos · Answer 1 · 2020-09-17T16:00:16+0000

If x' and x" be the x-coordinate of the particle at initial time t' and t" respectively then vx =(x"-x')/(t"-t') = tanθ.

For t"-t' infinitesimally small it is the vx at that instant.

So slope of the tangent at any point in the above graph gives vx .

At t=t1, tanθ is positive, so sign of vx is positive.

At t= t2 the slope of the curve is horizontal, so tanθ=0 → vx =0.

At t=t3 the slope of the curve is negative, so sign of vx is negative.

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