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Exercises

  • First identify : what is given and what is required. Establish relative order between given and required attribute.
  • Use differentiation method to get a higher order attribute in the following order : displacement (position vector) → velocity → acceleration.
  • Use integration method to get a lower order attribute in the following order : acceleration → velocity → displacement (position vector).
  • Since we are considering accelerated motion in one dimension, graphical representation of motion is valid. The interpretation of plot in terms of slope of the curve gives higher order attribute, whereas interpretation of plot in terms of area under the plot gives lower order attribute.

A particle of mass m moves on the x-axis. It starts from rest t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate time (0<t<1). Prove that magnitude of instantaneous acceleration during the motion can not be less than 4 m / s 2 .

In between the starting and end point, the particle may undergo any combination of acceleration and deceleration. According to question, we are required to know the minimum value of acceleration for any combination possible.

Let us check the magnitude of acceleration for the simplest combination and then we evaluate other complex possible scenarios. Now, the simplest scenario would be that the particle first accelerates and then decelerates for equal time and at the same rate to complete the motion.

Velocity – time plot

Velocity – time plot

In order to assess the acceleration for a linear motion , we would make use of the fact that area under v-t plot gives the displacement (= 1 m). Hence,

Δ OAB = 1 2 x v t = 1 2 x v x 1 = 1 m v max = 2 m / s

Thus, the maximum velocity during motion under this condition is 2 m/s. On the other hand, the acceleration for this motion is equal to the slope of the line,

a = tan θ = 2 0.5 = 4 m / s 2

Now let us complicate the situation, in which particle accelerates for shorter period and decelerates gently for a longer period (see figure below). This situation would result velocity being equal to 2 m/s and acceleration greater than 4 m / s 2 .

It is so because time period is fixed (1 s) and hence base of the triangle is fixed. Also as displacement in given time is 1 m. It means that area under the triangle is also fixed (1 m). As area is half of the product of base and height, it follows that the height of the triangle should remain same i.e 2 m/s. For this reason the graphical representations of possible variation of acceleration and deceleration may look like as shown in the figure here.

Velocity – time plot

Area under velocity – time plot gives displacement , whereas slope of the plot gives the acceleration.

Clearly, the slope of OA’ is greater than 4 m / s 2 .

We may argue that why to have a single combination of acceleration and deceleration? What if the particle undergoes two cycles of acceleration and deceleration? It is obvious that such consideration will again lead to similar analysis for symmetric and non-symmetric acceleration, which would be bounded by the value of time and area of the triangle.

Thus we conclude that the minimum value of acceleration is 4 m / s 2 .

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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