A Honda Civic travels in a straight line along a road. Its distance x from a sto

Question

A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t)= t2 t3, where = 1.44 m/s2 and = 4.70×102 m/s3.

Part A- Calculate the average velocity of the car for the time interval t=0 to t1 = 1.91 s. (answer will be in m/s)

Part B- Calculate the average velocity of the car for the time interval t=0 to t2 = 3.92 s. (answer will be in m/s)

Part C- Calculate the average velocity of the car for the time interval t1 = 1.91 s to t2 = 3.92 s (answer will be in m/s)

Explanation / Answer

x = 1.44t² - 0.047t³

dx/dt = 2.88t - 0.141t² = V = instantaneous velocity of Civic

To find the Vavg between time ranges given, must evaluate V at each end of range, and apply the Vavg formula:
Vavg = (V1+V2)/2

Part A
at t =0,          V1 = 2.88(0) - 0.141(0)² = 0
at t= 1.91 sec, V2 = 2.88(1.91) - 0.141(1.91)² = 4.986 m/s

Vavg = (V1+V2)/2 = (4.986 + 0)/2 = 4.986/2 = 2.4937.94 m/s….. ANS

Part B
at t =0,          V1 = 2.88(0) - 0.141(0)² = 0
at t= 3.92 sec, V2 = 2.88(3.92) - 0.141(3.92)² = 9.123 m/s

Vavg = (V1+V2)/2 = (9.123 + 0)/2 = 9.123/2 = 4.5615 m/s….. ANS

Part C
at t =1.91,       V1 = 2.88(1.91) - 0.141(1.91)² = 4.986 m/s
at t= 3.92 sec, V2 = 2.88(3.92) - 0.141(3.92)² = 9.123 m/s

Vavg = (V1+V2)/2 = (4.986 +9.123)/2 = 7.0545    m/s….. ANS

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