A car drives down a road in such a way that its velocity ( in m/s) at time t (seconds) is
v(t)=1t^(1/2)+1
.
Find the car's average velocity (in m/s) between t=2 and t=6. ?
I am lost!
If we consider the average velocity as the total distance travelled in a given time, then
V̄
= [∫(sqrt(t)+1)dt ] / (6-2)
= [(2/3)t^(3/2)+t] from 2 to 6 / (6-2)
= (15.80/-3.88)/4
=2.98 m/s
To find the car's average velocity between t=2 and t=6, we need to find the displacement during this time period and divide it by the time interval.
The displacement can be found by integrating the velocity function:
∫(v(t))dt = ∫(1t^(1/2) + 1)dt
= ∫(t^(1/2))dt + ∫(1)dt
Integrating each term separately:
= (2/3)(t^(3/2)) + t + C
where C is the constant of integration.
Now, to find the displacement between t=2 and t=6, we substitute these values into the above equation:
Displacement = [(2/3)(6^(3/2)) + 6] - [(2/3)(2^(3/2)) + 2]
= [(2/3)(6^(3/2)) + 6] - [(2/3)(2)(2^(1/2)) + 2]
= [(2/3)(6^(3/2)) + 6] - [(4/3)(2^(1/2)) + 2]
= [(2/3)(6^(3/2)) + 6] - [(4/3)(2^(1/2)) + 2]
= [(2/3)(36√6) + 6] - [(4/3)(2^(1/2)) + 2]
Next, we divide the displacement by the time interval:
Average velocity = Displacement / Time interval
= [(2/3)(36√6) + 6] - [(4/3)(2^(1/2)) + 2] / (6 - 2)
Simplifying the expression further will give us the average velocity.
No problem! I can help you with that.
To find the car's average velocity between t = 2 and t = 6, we need to calculate the total distance traveled during that time interval and divide it by the duration. The average velocity is defined as the total displacement divided by the time taken.
Here are the steps to find the car's average velocity:
Step 1: Calculate the displacement of the car during the time interval.
To do this, we need to find the distance traveled at t = 6 seconds and t = 2 seconds.
At t = 6 seconds, the velocity function gives us:
v(6) = 1(6)^(1/2) + 1
v(6) = 1(√6) + 1
v(6) = √6 + 1
Similarly, at t = 2 seconds, the velocity function gives us:
v(2) = 1(2)^(1/2) + 1
v(2) = 1(√2) + 1
v(2) = √2 + 1
The displacement, Δx, is given by:
Δx = x(6) - x(2)
where x(t) represents the position function. However, since we only have the velocity function, we need to approximate the displacement using the trapezoidal rule.
Δx ≈ Δt/2 * (v(6) + v(2))
where Δt is the time interval, which is 6 - 2 = 4 seconds.
Step 2: Calculate the average velocity.
The average velocity is given by:
Average velocity = Δx / Δt
Substituting the values we obtained earlier:
Average velocity ≈ Δt/2 * (v(6) + v(2)) / Δt
Average velocity ≈ 2 * (√6 + 1 + √2 + 1) / 4
Simplifying further:
Average velocity ≈ (√6 + √2 + 2) / 2
Therefore, the car's average velocity between t = 2 and t = 6 is approximately (√6 + √2 + 2) / 2 m/s.
V(t) = t^(1/2) + 1 = sqrt(t) + 1.
V(2) = 2^(1/2) + 1 + 2.41 m/s
V(6)= 6^(1/2) + 1 = 3.45 M/S
Vavg = (2.41 + 3.45) / 2 = 2.93 m/s