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 t^2− β t^3, where α = 1.54m/s2 and β = 5.05×10−2m/s^3 .
Calculate the average velocity of the car for the time intervals:
t=0 to t1 = 2.02s .
t=0 to t2 = 4.00s .
t1 = 2.02s to t2 = 4.00s
Thank you!!
avg velocity= (finalvelocity+initialvel)/2
first, find the stop and start velocityies
t(0)
t(2.02)
t(4.00)
I will do one for you, the last one.
avgvel=(v(4.00)+v(2.02))/2
To calculate the average velocity of the car for a given time interval, we need to find the displacement of the car during that interval and divide it by the total time taken.
For t = 0 to t1 = 2.02s:
1. We need to find the displacement of the car during this time interval. The displacement is the difference between the position at t1 and the position at t = 0.
x(t1) - x(0) = (α t1^2 - β t1^3) - (α (0)^2 - β (0)^3) = α t1^2 - β t1^3
2. The average velocity is given by the displacement divided by the total time taken.
Average velocity = (α t1^2 - β t1^3) / (t1 - 0) = (α t1^2 - β t1^3) / t1
Plug in the values of α, β, and t1 into the above equation to get the average velocity.
For t = 0 to t2 = 4.00s:
1. The displacement of the car during this time interval is:
x(t2) - x(0) = (α t2^2 - β t2^3) - (α (0)^2 - β (0)^3) = α t2^2 - β t2^3
2. The average velocity is:
Average velocity = (α t2^2 - β t2^3) / (t2 - 0) = (α t2^2 - β t2^3) / t2
Plug in the values of α, β, and t2 into the above equation to get the average velocity.
For t1 = 2.02s to t2 = 4.00s:
1. The displacement of the car during this time interval is:
x(t2) - x(t1) = (α t2^2 - β t2^3) - (α t1^2 - β t1^3) = α (t2^2 - t1^2) - β (t2^3 - t1^3)
2. The average velocity is:
Average velocity = (α (t2^2 - t1^2) - β (t2^3 - t1^3)) / (t2 - t1)
Plug in the values of α, β, t1, and t2 into the above equation to get the average velocity.
I hope this explanation helps you in calculating the average velocities for the given time intervals!