The speed of the belt is changed at a uniform rate from 0 to

The speed of the belt is changed at a uniform rate from 0 to 2 m/s during a time interval of 0.2 second. Calculate (a) the distance traveled by the belt during the 0.2-second interval; and (b) the maximum acceleration of any point on the belt during this interval.

Solution

solution:

(a)

Distance travelled = 2 x 0.2 = 0.4 m

(b) Uniform speed change a_t = dv/dt = 2/0.2 = 10m/s²

Maximum acceleration = ((a_r²) + a_t²)^1/2  =( 2²/0.2 + 10²)^1/2= 10.95 m/s²