3.4 Motion with constant acceleration  (Page 7/10)

Strategy

Solution

1. Equation for the gazelle: The gazelle has a constant velocity, which is its average velocity, since it is not accelerating. Therefore, we use [link] with $x_0=0$ :
   
   $x=x_0+\stackrel{\text{–}}{v}t=\stackrel{\text{–}}{v}t.$
   
   Equation for the cheetah: The cheetah is accelerating from rest, so we use [link] with $x_0=0$ and $v_0=0$ :
   
   $x=x_0+v_{0}t+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2.$
   
   Now we have an equation of motion for each animal with a common parameter, which can be eliminated to find the solution. In this case, we solve for t :
   
   $\begin{array}{}\\ \\ x=\stackrel{\text{–}}{v}t=\frac{1}{2}a{t}^2\hfill \\ t=\frac{2\stackrel{\text{–}}{v}}{a}.\hfill \end{array}$
   
   The gazelle has a constant velocity of 10 m/s, which is its average velocity. The acceleration of the cheetah is 4 m/s ² . Evaluating t , the time for the cheetah to reach the gazelle, we have
   
   $t=\frac{2\stackrel{\text{–}}{v}}{a}=\frac{2(10)}{4}=5\text{s}\text{.}$
2. To get the displacement, we use either the equation of motion for the cheetah or the gazelle, since they should both give the same answer.
   Displacement of the cheetah:
   
   $x=\frac{1}{2}a{t}^2=\frac{1}{2}(4){(5)}^2=50\text{m}\text{.}$
   
   Displacement of the gazelle:
   
   $x=\stackrel{\text{–}}{v}t=10(5)=50\text{m}\text{.}$
   
   We see that both displacements are equal, as expected.

Significance

Summary

• When analyzing one-dimensional motion with constant acceleration, identify the known quantities and choose the appropriate equations to solve for the unknowns. Either one or two of the kinematic equations are needed to solve for the unknowns, depending on the known and unknown quantities.
• Two-body pursuit problems always require two equations to be solved simultaneously for the unknowns.

Conceptual questions