A cyslist is traveilling on a circular section of highway of radius 2500 ft at the speed of 60 mile/h. The cyclist suddenly applies the brakes causing the bicycle to slow down at constant rate. Knowing that after 8 second,the speed has been reduced to 45 mile/h. The acceleration of the bicyle immediately after the breakes have been applied. is

Options

A. `2ft//s^(2)`
B. `4.14ft//s^(2)`
C. `3.10ft//s^(2)`
D. `2.75 ft//s^(2)`

Answer

B

Explanation

Tangential acceleration
` = a_(t)=(v_(f)-v_(i))/(t)`
` = ((45-60)/(8)) ((22)/(15)) = -(11)/(4)ft//s^(2)`
The radial acceleration is
`a = (v_(2))/(r) = ((60xx(22)/(15))^(2))/(200)`
` = 3.1ft//s^(2)`
`therefore a = sqrt(a_(r)^(2) +a_(t)^(2))`
or`a = 4.14ft//s^(2)`