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a 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared east for 30 seconds. what is the speed
of the cart at the end of this 3.0 second interval?

V = 20.5 m/s

Explanation:

Given,

The mass of the cart, m = 6 Kg

The initial speed of the cart, u = 4 m/s

The acceleration of the cart, a = 0.5 m/s²

The time interval of the cart, t = 30 s

The final velocity of the cart is given by the first equation of motion

                              v = u + at

                                  = 4 + (0.5 x 30)

                                = 19 m/s

Hence the final velocity of cart at 30 seconds is, v = 19 m/s

The speed of the cart at the end of  3 seconds

                                    V = 19 + (0.5 x 3)

                                       = 20.5 m/s

Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s

d) physical means can be used to separate the components of air.