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The parachute on a race car of weight 8,830 N opens at the end of a quarter-mile run when the car...

Question:

The parachute on a race car of weight {eq}8,830 \ N {/eq} opens at the end of a quarter-mile run when the car is traveling at {eq}38 \ m / s {/eq}. What total retarding force must be supplied by the parachute to stop the car in a distance of {eq}1,055 \ m {/eq}?

Kinematics Problems:

In this problem, we can use the kinematic equation in the form;

{eq}v_f^2 = v_i^2 + 2 aD {/eq}

where {eq}v_i {/eq} is the initial velocity, {eq}v_f {/eq} is the final velocity, {eq}a {/eq} is the acceleration and {eq}D {/eq} is the distance traveled.

Answer and Explanation: 1


Given:

  • Weight of Racecar ({eq}W_c {/eq}) = 8830 N
  • Initial velocity of racecar ({eq}v_i {/eq}) = 38 m/s
  • Final velocity of racecar ({eq}v_f {/eq}) = 0 m/s
  • Stopping distance ({eq}D {/eq}) = 1055 m


Plugging in the values to solve for the acceleration;

{eq}\begin{align} (0)^2 &= (38)^2 + 2 a(1055)\\[0.25cm] 0 &= 1444 + 2110a\\[0.25cm] a &= -\frac{1444}{2110}\\[0.25cm] a &= -\frac{722}{1055}\ \rm m/s^2 \end{align} {/eq}


To find the retardation force, we use the formula for force as;

{eq}\begin{align} F &= ma\\[0.25cm] F &= \left(\frac{W_c}{g}\right)a\\[0.25cm] F &= \frac{(8830)}{(9.81)}\left(-\frac{722}{1055}\right)\\[0.25cm] F &\approx \boxed{-616\ \rm N} \end{align} {/eq}

The negative sign just signifies the direction of the force which is the opposite of the motion.


Learn more about this topic:

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Solving Kinematics Problems

from

Chapter 2 / Lesson 21
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How do we use physics to answer questions about motion? One way is with kinematics, and in this lesson we'll examine several kinematics equations and how to use them.


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