Total Accel: (15)^2/173 + .6614 m/s^2

[runner1738]

at the moment the train's speed is 54 km/h. what is the magnitude of the total acceleration? units m/s^2 r=173 and it takes 18.9 s to slow down from 82 km/h to 37 km/h, the a tangential is .6614 so is it (15)^2/173 + .6614 = total?

 

[Andrew Mason]

at the moment the train's speed is 54 km/h. what is the magnitude of the total acceleration? units m/s^2 r=173 and it takes 18.9 s to slow down from 82 km/h to 37 km/h, the a tangential is .6614 so is it (15)^2/173 + .6614 = total?

From your question it appears that we have a train on a curved track with radius 173 m. and slowing down at the rate of 12.5 m/sec in 18.9 seconds. You want to know its total acceleration when its speed is 54 km/hr (15 m/s).

In order to work this out, you have to add them as vectors. The accelerations are perpendicular to each other. So:

$$a_l^2 + a_c^2 = a_{total}^2$$

AM

 

[wais]

Yes, that is correct. The magnitude of the total acceleration can be calculated by using the formula: a = (v2 - v1)/t, where v2 is the final speed (37 km/h), v1 is the initial speed (82 km/h), and t is the time taken (18.9 seconds). Plugging in the values, we get a = (37 km/h - 82 km/h)/(18.9 s) = -2.38 m/s^2. This is the tangential acceleration. To calculate the total acceleration, we need to add the centripetal acceleration, which is given by a = (15)^2/173 = 1.31 m/s^2. Therefore, the magnitude of the total acceleration is 1.31 m/s^2 + 2.38 m/s^2 = 3.69 m/s^2. It is important to note that the units for the total acceleration should be in m/s^2, as given in the question.