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horsena [70]
3 years ago
15

A car traveling initially at 9.49 m/s accelerates at the rate of 0.988 m/s^2 for 3.05s. What is it’s velocity at the end of the

acceleration?
Physics
2 answers:
krok68 [10]3 years ago
7 0

Answer:

• From first equation of motion:

{  \boxed{ \bf{v = u + at}}}

substitute:

v = 9.49 + (0.988 \times 3.05) \\ v = 9.49 + 3.0134 \\ v = 12.50 \:  \: m {s}^{ - 1}

weeeeeb [17]3 years ago
6 0

Answer:

12.50 m/s

Explanation:

Vi = 9.49 m/s

a = 0.988 m/s²

t = 3.05 s

Vf = ?

Vf = Vi + at

Vf = 9.49 + (0.988)(3.05)

Vf = 12.50 m/s

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Answer:

t = 0.33h = 1200s

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Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

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car 2:

x'=v_2t    (2)

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For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}

t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s

The position in which both cars coincides is:

x=(55km/h)(0.33h)=18.33km

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