Velocity-Time Graphs & Acceleration Video Lessons

Concept

# Problem: Figure P2.17 shows a graph of vx versus t for the motion of a motorcyclist as he starts from rest and moves along the road in a straight line. (a) Find the average acceleration for the time interval t = 0 to t = 6.00 s. (b) Estimate the time at which the acceleration has its greatest positive value and the value of the acceleration at that instant.(c) When is the acceleration zero? (d) Estimate the maximum negative value of the acceleration and the time at which it occurs.

###### FREE Expert Solution

Acceleration is expressed as:

$\overline{){\mathbf{a}}{\mathbf{=}}\frac{\mathbf{∆}\mathbf{v}}{\mathbf{∆}\mathbf{t}}}$, where Δv is the change in velocity and Δt is the time interval over this change.

Δv = vf - vi

Similarly, Δt = tf - ti

So that we now have:

$\overline{){\mathbf{a}}{\mathbf{=}}\frac{{\mathbf{v}}_{\mathbf{f}}\mathbf{-}{\mathbf{v}}_{\mathbf{i}}}{{\mathbf{t}}_{\mathbf{f}}\mathbf{-}{\mathbf{t}}_{\mathbf{i}}}}$

(a)

The acceleration of the velocity-time graph is the slope of the graph.

The average acceleration for the time interval t = 0 to t = 6.00 s is:

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###### Problem Details

Figure P2.17 shows a graph of vx versus t for the motion of a motorcyclist as he starts from rest and moves along the road in a straight line.

(a) Find the average acceleration for the time interval t = 0 to t = 6.00 s.

(b) Estimate the time at which the acceleration has its greatest positive value and the value of the acceleration at that instant.

(c) When is the acceleration zero?

(d) Estimate the maximum negative value of the acceleration and the time at which it occurs.