Respuesta :
Answer:
Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)
Explanation:
Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.
[tex]X(van)=5.65t+154[/tex]
[tex]X(driver)=34.4t+\frac{(-2)t^{2} }{2}[/tex]
or by rearanging the drivers equation.
[tex]X(driver)=34.4t+t^{2}[/tex]
Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.
[tex]X(van)=X(driver)[/tex]
[tex]5.65t+154=34.4t-t^{2}[/tex]
[tex]0=t^{2} -(34.4-5.65)t+154[/tex][tex]0=t^{2} -28.75t+154[/tex]
To solve this equation we use the following formulas
[tex]t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}[/tex]
Where a=1; b=-28.75; c=154
So we get:
[tex]t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s[/tex][tex]t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s[/tex]
At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.
[tex]V(driver)=V_{0} +at[/tex]}
[tex]V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}[/tex][tex]V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}[/tex]
This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).
Best of luck
