Question

A train slows down as it rounds a sharp horizontal turn, going from 92.0 km/h to 54.0 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 54.0 km/h. Assume the train continues to slow down at this time at the same rate.

Answer 1

Calculation of centripetal acceleration(ac):

`a_c=(v^2)/(r)`

we can change our speed in terms of m/s

`v=54km/h=54*(1000)/(3600) m/s`

now, we can plug these values

`a_c=((54*(1000)/(3600))^2)/(130)`

`a_c=1.73077m/s^2`

Calculation of tangential acceleration(at):

`a_t=(\Delta v)/(\Delta t)`

now, we can plug values

`a_t=((54-92)*(1000)/(3600) )/(16)`

now, we can simplify it

`a_t=-0.65972 m/s^2`

Calculation of resultant acceleration:

`a=√((a_c)^2+(a_t)^2)`

now, we can plug values

`a=√((1.73077)^2+(-0.65972)^2)`

`a=1.85224m/s^2`.............Answer