Solved

 A petrol car is parked 35 feet from a long warehouse (see figure). The revolving \text { A petrol car is parked } 35 \text { feet from a long warehouse (see figure). The revolving }

Question 76

Multiple Choice

 A petrol car is parked 35 feet from a long warehouse (see figure) . The revolving \text { A petrol car is parked } 35 \text { feet from a long warehouse (see figure) . The revolving } light on top of the car turns at a rate of 12\frac { 1 } { 2 } revolution per second. The rate at which the light beam moves along the wall is r=35πsec2θft/secr = 35 \pi \mathrm { sec } ^ { 2 } \theta \mathrm { ft } / \mathrm { sec } . Find the rate rr when θ\theta is π6\frac { \pi } { 6 } .
\text { A petrol car is parked } 35 \text { feet from a long warehouse (see figure) . The revolving }  light on top of the car turns at a rate of  \frac { 1 } { 2 }  revolution per second. The rate at which the light beam moves along the wall is  r = 35 \pi \mathrm { sec } ^ { 2 } \theta \mathrm { ft } / \mathrm { sec } . Find the rate  r  when  \theta  is  \frac { \pi } { 6 } .    A)   r = \frac { 140 } { 3 } \mathrm { tt } / \mathrm { sec }  B)   r = \frac { 70 \sqrt { 3 } \pi } { 3 } \mathrm { ft } / \mathrm { sec }  C)   r = \frac { 70 \sqrt { 3 } } { 3 } \mathrm { ft } / \mathrm { sec }  D)  r = \frac { 140 \pi } { 3 } \mathrm { ft } / \mathrm { sec }  E)  r = \frac { 70 \pi } { 3 } \mathrm { ft } / \mathrm { sec }


A) r=1403tt/secr = \frac { 140 } { 3 } \mathrm { tt } / \mathrm { sec }
B) r=703π3ft/secr = \frac { 70 \sqrt { 3 } \pi } { 3 } \mathrm { ft } / \mathrm { sec }
C) r=7033ft/secr = \frac { 70 \sqrt { 3 } } { 3 } \mathrm { ft } / \mathrm { sec }
D) r=140π3ft/secr = \frac { 140 \pi } { 3 } \mathrm { ft } / \mathrm { sec }
E) r=70π3ft/secr = \frac { 70 \pi } { 3 } \mathrm { ft } / \mathrm { sec }

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions