Answer to Question #277194 in Physics for Fahad

Question

2. At a sharp horizontal circular turn a train slows down going from 90.0 km/h to 50.0 km/h in the 15.0 s it takes to round the bend. The radius of the curve is 150 m. Assume the train continues to slow down at the same rate. At the moment the train speed reaches 50.0 km/h calculate:

(a) (3 marks) the magnitude and direction of the tangential acceleration.

(b) (3 marks) the radial acceleration (magnitude and direction).

(c) (4 marks) the magnitude and direction of the total acceleration.

Answer 1

$90\ km/h=25\ m/s$

$50\ km/h=13.9\ m/s$

(a) $v=v_0+a_\tau t\to a_\tau=(v-v_0)/t=(13.9-25)/15=-0.74\ (m/s^2)$ . Tangential acceleration is directed along the tangent to the trajectory and opposite to the direction of the velocity.

(b) $a_n=v^2/R=13.9^2/150=1.29\ (m/s^2)$ . The radial acceleration is directed to the center of curvature of the trajectory.

(c) $a=\sqrt{0.74^2+1.29^2}=1.49\ (m/s^2)$

$\theta=\tan^{-1}\frac{a_n}{a_\tau}=\tan^{-1}\frac{1.49}{0.74}=63.6°$ relative to the direction of tangential acceleration.