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3 years ago

6

A car's speed changes from 10m/s to 35m/s in 15.0s along a straight road while moving in one way. Find the magnitude of the acce

leration of the car during this period.

1 answer:

Vinil7 [7]3 years ago

6 0

Answer:

1.67 m/s²

Explanation:

Applying,

a = (v-u)/t................ Equation 1

Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

From the question,

Given: v = 35 m/s, u = 10 m/s, t = 15.0 s

Substitute these values into equation 1

a = (35-10)/15

a = 25/15

a = 1.67 m/s²

Hence the acceleration of the car is 1.67 m/s²

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$\sum F = m_A a$

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Replacing the values previously found,

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

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

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

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

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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,

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

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

$\%error = 2.7%$

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3 years ago