Question

An electron moving along the x axis has a position given by x = 10te^minus3t m, where t is in seconds. How far is the electron from the origin when it momentarily stops? m A car traveling 56.0 km/h is 20.0 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.39 s later. What is the magnitude of the car's constant acceleration before impact? m/s^2 How fast is the car traveling at impact? m/s

Answer 1

7.- We have that the position of the electron is given by

$x=10te^{-3t}$

Then the velocity of the electron is given by the derivate of the position function

$v=\frac{dx}{dt}$

$v=\frac{d}{dt}\left(10te^{-3t} \right )$ ......................By the derivate of the product

$v=10e^{-3t}+(-3)10te^{-3t}$

$v=(1-3t)10e^{-3t}$

Then we want to know when the velocity is zero

$v=0$

$0=(1-3t)10e^{-3t}$ ......only can be zero becasue the $e^{-3t}$ is allways different to zero

$0=(1-3t)$

$3t=1$

$t=\frac{1}{3}$ .........................The unit of must be $\rm s^{-1}$ , becaues the argument of $e^{()}$ cannot have unit

$t=\frac{1}{3}\rm\;s$

Replacinf that time in the equation of the position

$x=10\left(\frac{1}{3}\rm\;s \right )e^{-3(1/3\rm \;s)}$ ......The unit of must be $\rm \;m/s$

$x=1.23\rm\;m$

8.- The intial velocitu in m/s is

$v_o =56.0\rm \; km/h \left(\frac{1000\rm\;m}{1\rm\;km} \right )\left(\frac{1\rm\;h}{3600\rm\;s} \right )$

$v_o=15.6\rm\;m/s$

a) Here we go to ude the equation of the position in a motion with constant acceleration

$\Delta x= v_ot+\frac{1}{2}at^2$ ...from this equation

$\frac{1}{2}at^2=\Delta x- v_ot$

$a=\frac{2(\Delta x- v_ot)}{t^2}$

$a=\frac{2[(20\rm\;m)- (15.6\rm\;m/s )(2.39\; s)]}{(2.39\; \rm s)^2}$

$a=-6.05\rm\;m/s^2$

The acceleration is negative because is deceleration. But its magnitude is

$|a|=6.05\rm\;m/s^2$

b)Now we have the acceleration of the motion. Then we can find the final velocity is given by the equation

$v_f=v_o+at$

$v_f=15.6\rm\;m/s+(-6.05\rm\;m/s^2)(2.39\rm \;s)$

$v_f=1.14\rm\;m/s$