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torisob [31]
3 years ago
6

A sport car moving at constant speed travels 110m in 5.0 s. if it then brakes and comes to a stop in 4.0 s, what is the magnitud

e of its acceleration in m/s2, and in g’s (g=9.80 m/s2)? 2
Physics
1 answer:
stiks02 [169]3 years ago
7 0
D = 110 m,  t = 5 s
v o = 110 cs : 5 m = 22 m/s
-------------------------------------
v = v o - a t
v = 0 m/s,  v o =  22 m/s,  t = 4 s
0 = 22 - 4 a
4 a = 22
a = 22 : 4
a = 5.5 m/s²
g = 9.80 m/s²
9.80 : 5.5 = 0.56 
Answer:
The magnitude of its acceleration is 5.5 m/s or 0.56 g. 
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To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

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\sum F = m_b a

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Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

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a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
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mojhsa [17]

Answer:

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HOPE THIS HELPED:)

8 0
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