A train slows down a t a constant rate as it rounds a sharp circular horizontal

Question

A train slows down a t a constant rate as it rounds a sharp circular horizontal turn. Its initial speed is not known. It takes 19.3 s to slow down from 86 km/h to 29 km/h. The radius of the curve is 189 m. As the train goes around the turn, what is the magnitude of the tangential component of the acceleration? Answer in units of m/s2. As the train goes around the turn, what is the sign of the tangential component of the acceleration? Take the direction in which the train is moving to be the positive direction. Not enough information + - 0 At the moment the train's speed is 54 km/h, what is the magnitude of the total acceleration? Answer in units of m/s2.

Explanation / Answer

1 ) Tangential acceleration   a_t    = - d v /dt                                   = - ( 86 - 29 km/h ) / 19.3 s                                    = - (23.8 m/s - 8.0m/s ) / 19.3 s    ( 1 km/h = 0.2778 m/s )                                     = - 0.81 m/s^2 Radial acceleration is a_r = v^2 / r               speed at that moment is 54 km/h                                                v = 15.0 m/s                 radius of the curve is r = 189 m                      a _r   = ( 15.0 m/s )^2 / 189m                               = 1.19 m/s^2 Total acceleration is a = v a_t ^2 +a _r ^2                                      = v ( - 0.81 )^2 + ( 1.19 )^2                                       = 1.43 m/s^2        direction is negative

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