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A car enters a freeway with initial velocity of 15.0 m/s and with con stant rate of acceleration, reaches a velocity of 22.5 m/s in a time interval of 3.50 s. a) Determine the value of the car's acceleration. b) Determine the distance traveled by the car in this 3.50 s time interval. c) Determine the average velocity of the car over this 3.50 s time interval.

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Answer:

a) The acceleration is 2.14 m/s^{2}

b) The distance traveled by the car is 65.61 m

c) The average velocity is 18.75 m/s

Step-by-step explanation:

Using the equations that describe an uniformly accelerated motion:

a)
a=(v_f - v_o)/(t) =(22.5m/s - 15.0 m/s)/(3.50s)

b)
d= d_0 + v_0 t + (1)/(2) a t^(2) = 0 +15.0 x 3.5 + (2.14x3.50^(2) )/(2) = 65.61 m

c)
v_m =(d)/(t)=(65.61)/(3.5)  =18.75 m/s

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