A train slows down as it rounds a sharp horizontal turn, going from 84.0 km/h to

Question

A train slows down as it rounds a sharp horizontal turn, going from 84.0 km/h to 42.0 km/h in the 16.0 s that it takes to round the bend. The radius of the curve is 130 m. Compute the acceleration at the moment the train speed reaches 42.0 km/h. Assume the train continues to slow down at this time at the same rate. magnitude .729 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2 direction ° backward (behind the radial line pointing inward)

Explanation / Answer

here

v1 = 84 km/hr = 23.34 m/s

v2 = 42km/hr = 11.67 m/s

ac = v2^2 / r

ac = 11.67^2 / 130

ac = 1.05 m/s^2

then at tengential

at = delta V / delta t

at = ( 42 - 84 ) * 1000/3600 / 16

at = -0.729 m/s^2

then

a = sqrt( ac^2 + at^2)

a = sqrt( 1.05^2 + 0.729^2)

a = 1.278 m/s^2

and then the direction is

theta = tan^-1( at / ac)

theta = tan^-1( 0.729 / 1.05)

theta = 34.77 deg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.