Solved Problem on Kinematics
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A motorcyclist is moving in the opposite direction of a reference frame, the magnitude of its initial speed is 25 m/s, at the initial time its position is -150 m, and the magnitude of deceleration is 2 m/s2. Find:
a) The equation of displacement as a function of time;
b) The equation of velocity as a function of time;
c) The instant in which it passes through the origin of the reference frame;
d) The instant that its speed is zero.

Problem data
• the initial speed of the rider:    |v0| = 25 m/s;
• acceleration of the motorcyclist:    |a| = 2 m/s2;
• position in the initial instant:    S0 = −150 m.
Problem diagram.

We chose a reference frame oriented to the right.

figure 1

Solution

a) The equation of displacement as a function of time is of the type
$\bbox[#99CCFF,10px] {S=S_{0}+v_{0}t+\frac{a}{2}\;t^{2}}$
The initial position is already given in the problem, S0 = −150 m as the motorcyclist moves in the opposite direction to the orientation of the reference frame its speed is negative (v<0) so we have v = −25 m/s. The motorcyclist is decelerating, its acceleration is in the opposite direction of speed (a>0), thus a= 2 m/s2. The displacement equation as a function of time
$S=-150-25t+\frac{2}{2}\;t^{2}$
$\bbox[#FFCCCC,10px] {S=-150-25t+t^{2}}$

b) For the equation of velocity as a function of time is of type
$\bbox[#99CCFF,10px] {v=v_{0}+at}$
with the problem data, we have

$\bbox[#FFCCCC,10px] {v=-25t+2t}$

c) When the motorcyclist passes through the origin we have S = 0, substituting this value in the expression found in item (a), we have.
$0=-150-25\;t+t^{2}$
This is a Quadratic Equation where the unknown is the value of time.

the solution of the Quadratic Equation    $$t^{2}-25\;t-150=0$$
$\Delta =b^{2}-4ac=(\;-25\;)^{2}-4 \times 1 \times (\;-150\;)=625+600=1225\\ {}\\ t=\frac{-b\pm \sqrt{\;\Delta \;}}{2a}=\frac{-(\;-25\;)\pm \sqrt{\;1225\;}}{2 \times 1}=\frac{25\pm 35}{2}$
the two roots of the equation will be
$$t_{1}=30\;\text{s}$$       and       $$t_{2}=-5\;\text{s}$$

Since there is no negative time, we neglected the second root, it will go through the origin at t = 30 s (it starts the motion to the left of the origin, its speed decreases due to deceleration until it becomes zero, then changes direction and begins to move in the same direction of the reference frame until it passes through the origin).

d) When the speed of the motorcycle cancels we have v = 0, substituting this value in the expression found in item (b), we have
$0=-25+2t$
This is a Linear Equation where the unknown is the value of time
$2t=25\\ t=\frac{25}{2}$
$\bbox[#FFCCCC,10px] {t=12.5\;\text{s}}$
This is the instant that the motorcyclist changes direction and begins to move in the same direction of the referential frame until passing through the origin.