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3 years ago

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A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.

Then it slows down at a steady rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?

1 answer:

wlad13 [49]3 years ago

5 0

Answer:

12m/s

Explanation:

$v_f=v_o+at$

Let's call the velocity that the car maintains for 10 seconds $v_f_1$ , and the final velocity $v_f_2$ .

$v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s$

Hope this helps!

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Generally we can easily solving mathematically substitute into v_1

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$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

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Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

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