Question

A train slows down as it rounds a sharp horizontal turn, going from 98.0 km/h to 52.0 km/h in the 16.0 s that it takes to round the bend. The radius of the curve is 140 m. Compute the acceleration at the moment the train speed reaches 52.0 km/h. Assume the train continues to slow down at this time at the same rate.

magnitude	m/s2
direction	° backward (behind the radial line pointing inward)

Answer 1

here,

initial speed , u = 98 km/h = 10.3 m/s

final speed , v = 52 km/h = 5.44 m/s

time interval , t = 16 s

the tangential accelration be at

v = u + a * t

5.44 = 10.3 + a * 16

solving for at

at = - 0.3 m/s^2

for radial accelration ,

speed , v1 = 52 km/h = 14.4 m/s

radius , r = 140 m

the radial accelration , ar = v1^2 /r

ar = 14.4^2 /140 m/s^2

ar = 1.49 m/s^2

the magnitude of accelration , a= sqrt(at^2 + ar^2)

a = sqrt(1.49^2+ 0.3^2)

a = 1.52 m/s^2

the direction , theta = arctan(at/ar)

theta = arctan(- 0.3/1.49) = - 11.4 degree

theta = 11.4 degree backward