v\left(t\right)=5-tv(t)=5−t represents the velocity of an object over the interval [0, 7] in meters per second. What is the displacement of the object over the time interval?
To find the displacement of the object over the time interval [0, 7], we can use the definite integral of the velocity function. The definite integral of a function represents the area under the curve of that function over a given interval.
In this case, the velocity function is v(t) = 5 - t. To find the displacement, we need to evaluate the definite integral of this velocity function over the interval [0, 7].
∫[0,7] (5 - t) dt
To evaluate this definite integral, we can use the power rule of integration. The power rule states that if the function is of the form f(x) = ax^n, where a and n are constants, then the indefinite integral is F(x) = (a/n+1)x^(n+1) + C. In this case, since we have v(t) = 5 - t, we can integrate each term separately:
∫[0,7] 5 dt = 5t | [0, 7] = 5(7) - 5(0) = 35
∫[0,7] -t dt = -t^2/2 | [0, 7] = -(7^2)/2 - (0^2)/2 = -49/2
Now, we can substitute the bounds of the integral to find the displacement:
Displacement = 35 - (-49/2) = 35 + 49/2 = 35 + 24.5 = 59.5 meters
Therefore, the displacement of the object over the time interval [0, 7] is 59.5 meters.