Answer to Question #278241 in Mechanics | Relativity for Mishita

Question

2. At a sharp horizontal circular tum 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)the magnitude and direction of the tangential acceleration.

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

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

Answer 1

Explanations & Calculations

a)

• Tangential acceleration is what the train reduces speed linearly. This can be calculated using $v=u+at$ as $\small v=50kmh^{-1}$ , $\small u =90kmh^{-1}$ and $\small t = 15s$ . make sure you convert the speeds into $\small ms^{-1}$ before substituting to that equation.
• The direction is along the instantaneous direction of motion.

b)

• Radial acceleration is along the radius of the turn and directs away from the train to the centre of the turn.
• Once the tangential acceleration and the radius of the turn are known, the needed quantity can be calculated using the equation $\small a =r\alpha$ in which $a \,\,\&\,\, \alpha$ represents the linear and tangential accelerations respectively.

c)

• Both accelerations are at 90 degrees to each other. You have calculated both and you know the magnitudes as well.
• By using the Pythagorean equation you can find the magnitude of the total acceleration and using tangent identity for the triangle formed by those 3 accelerations, the direction of the total acceleration can be found.

• Let me know once you have given a try