Calculate the average velocity of the car for the time interval t=0 to t1 = 1.91s?
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)= α β t3, where α = 1.43m/s2 and β =
1.) Calculate the average velocity of the car for the time interval t=0 to t1 = 1.91s.
2.) Calculate the average velocity of the car for the time interval t=0 to t2 = 3.93s.
3.) Calculate the average velocity of the car for the time interval t1 = 1.91s to t2 = 3.93s.
Could someone show me how to do this step by step? Completely lost
To determine the average velocity, use the following equation.
V ave = * (vi + vf)
x = 1.43 * t^2 - 4.5 * 10^-2 * t^3
To determine the equation for velocity versus time, take the first derivative of the equation above.
vf = vi + 2.86 * t - 0.135 * t^2
For the first 1.91 seconds, vi is 0 m/s.
vf = 2.86 * 1.91 - 0.135 * 1.91^2 = 4.9701065 m/s
V ave = * 4.9701065 = 2.48503525 m/s
For the second time period, vi is 0 and t is 3.93 seconds
vf = 2.86 * 3.93 - 0.135 * 3.93^2 = 9.1547385 m/s
V ave = * 9.1547385 = 4.57736925 m/s
For the third time period, time is not 3.93 seconds. To determine the time, subtract 1.91 seconds from 3.93, t = 2.02 seconds. Vi is 4.9701065 m/s.
vf = 4.9701065 + 2.86 * 2.02 - 0.135 * 2.02^2 = 10.1964525 m/s
V ave = * (4.57736925 + 10.1964525) = 7.386910875 m/s
I suggest that you check my calculations to determine if I have made any mistakes!
X(t) = position vector. Change in position over time is average velocity
1) [x(1.91) - x(0)] / [1.91 - 0]
2) [x(3.93) - x(0)] / [3.93 - 0]
3) [x(3.93) - x(1.91) / [3.93 - 1.91]
X(t) = α β where α = 1.43 and β =
= displacement / time, where displacement = x(tii) - x(ti) in a time interval (tii - ti)
1.) the average velocity of the car for the time interval t=0 to ti = 1.91s.
x(0) = 0
x(1.91) = α β = 1.43(3.6481) - (0.045)(6.967871) = 5.216783 - 0.313554195 = 4.903228805
=> = 4.903228805 / (1.91 - 0) = 2.5671355 m/s
2.) Calculate the average velocity of the car for the time interval t=0 to t2 = 3.93s.
x(0) = 0
x(3.93) = α β = 1.43(15.4449) - (0.045)(60.698457) = 22.086207 - 2.731430565 = 19.354776435
=> = 19.354776435 / (3.93 - 0) = 4.9248795 m/s
3.) the average velocity of the car for the time interval t1 = 1.91s to t2 = 3.93s
x(1.91) =4.903228805 m
x(3.93) = α β = 1.43(15.4449) - (0.045)(60.698457) = 22.086207 - 2.731430565 = 19.354776435 m
=> = [19.354776435 - 4.903228805] / (3.93 - 1.91) = 14.45154763 / 2.02 m/s = 7.1542315 m/s
hope this helps
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