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natka813 [3]
2 years ago
5

The state of illinois cycle rider safety program requires motorcycle riders to be able to brake from 30 mph (44 feet/second) to

zero in 45 feet. what constant deceleration is required to do that?
Physics
1 answer:
-Dominant- [34]2 years ago
6 0
<span>21.5 ft/s^2 The formula for distance traveled under constant acceleration is d = 0.5 AT^2 where d = distance A = acceleration T = time Solving for A, gives d = 0.5 AT^2 2d = AT^2 (1) 2d/T^2 = A The formula for velocity under constant acceleration is V = AT Solving for A, gives V = AT (2) V/T = A Now setting equations (1) and (2) above equal to each other and solving for T: V/T = 2d/T^2 TV = 2d (3) T = 2d/V Now substitute equation (3) into (2) above. V/T = A V/(2d/V) = A V * V/2d = A V^2/2d = A Plug in values and calculate. V^2/2d = A (44 ft/s)^2/(2*45 ft) = A (1936 ft^2/s^2)/(90 ft) = A 21.51111111 ft/s^2 = A Rounding to 3 significant figures gives 21.5 ft/s^2</span>
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Sedbober [7]

Answer:

a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa

Explanation:

a. Internal energy and the relative specific volume at s_1 are determined  from A-17:u_1=214.07kJ/kg, \ \alpha_r_1=621.2.

The relative specific volume at s_2 is calculated from the compression ratio:

\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825

#from this, the temperature and enthalpy at state 2,s_2 can be determined using interpolations T_2=862K and h_2=890.9kJ/kg. The specific volume at s_1 can then be determined as:

\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg

Specific volume,s_2:

\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg

The pressures at s_2 \ and\  s_3 is:

P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa

.The thermal efficiency=> maximum temperature at s_3 can be obtained from the expansion work at constant pressure during s_2-s_3

\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K

b.Relative SV and enthalpy  at s_3 are obtained for the given temperature with interpolation with data from A-17 :a_r_3=4.553 \ and\  h_3=1909.62kJ/kg

Relative SV at s_4 is

a_r_4=\frac{r}{r_c}\alpha _r_3

==\frac{16}{2}\times4.533\\=36.424

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa

Hence, the mean effective relative pressure is 674.95kPa

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