The velocity of a particle moving along the x-axis is given by f(t)=6-2t cm/sec. Use a graph of f(t)to find the exact change in position of the particle from time t=0 to t=4 seconds.
Oh, I see you're playing graph detective! Let's put on our Sherlock Holmes hats and solve this mystery.
To find the exact change in position of the particle from time t=0 to t=4 seconds, we need to calculate the area under the graph of f(t) between these two values.
Since the graph is a straight line with a negative slope of -2, it means the velocity is decreasing over time. So, the graph will look like a ramp leading downwards.
Now, let's calculate the area under the graph. We know that the area of a triangle is one-half the base times the height. In this case, the base is 4 seconds (from t=0 to t=4) and the height is 6 cm/sec (the initial velocity).
Using the formula: area = (1/2) * base * height, we can calculate the area:
area = (1/2) * 4 * 6 = 12 cm.
And there you have it! The exact change in position of the particle from t=0 to t=4 seconds is 12 cm.
Now, be careful not to trip over any math equations while investigating other physics problems! *Tip-toes away*
To find the exact change in position of the particle from time t=0 to t=4 seconds, we need to integrate the given velocity function, f(t), over the interval [0, 4].
The position function, F(t), can be found by integrating the velocity function with respect to time:
F(t) = ∫(f(t)) dt
Let's integrate the function f(t) = 6 - 2t with respect to t:
F(t) = ∫(6 - 2t) dt
To integrate, we can split the integral into two parts and apply the power rule for integration:
F(t) = ∫6 dt - ∫2t dt
Integrating each term separately:
F(t) = 6t - (2/2)t^2 + C
Where C is the constant of integration.
Now, we can find the change in position of the particle by evaluating the position function F(t) at t=4 and t=0:
Change in position = F(4) - F(0)
Substituting t=4 into the position function:
F(4) = 6(4) - (2/2)(4)^2 + C
F(4) = 24 - 32 + C
Substituting t=0 into the position function:
F(0) = 6(0) - (2/2)(0)^2 + C
F(0) = 0 + C
The constant of integration cancels out when finding the change in position, so we can ignore it.
Change in position = F(4) - F(0)
Change in position = (24 - 32) - 0
Change in position = -8 cm
Therefore, the exact change in position of the particle from t=0 to t=4 seconds is -8 cm.
To find the change in position of the particle from time t=0 to t=4 seconds, you can use the formula for displacement.
Displacement (Δx) is equal to the integral of the velocity function (f(t)) with respect to time (t) over the interval [0, 4]:
Δx = ∫[0,4] f(t) dt
Given that the velocity function is f(t) = 6 - 2t cm/sec, we can substitute this into the integral:
Δx = ∫[0,4] (6 - 2t) dt
To integrate this, we can apply the power rule of integration. The integral of a constant multiplied by t raised to the power of 1 is given by:
∫(a * t^n) dt = (a/n+1) * t^(n+1) + C,
where "a" is the constant coefficient, "n" is the exponent, and "C" is the constant of integration.
In our case, the coefficient "a" is -2 and the exponent "n" is 1. So, the integral becomes:
Δx = [-2/(1+1)] * t^(1+1) + C
Simplifying:
Δx = -t^2 + C
To find the constant of integration (C), we need to evaluate the displacement at t=0. Since the particle starts at t=0, the initial position is given by x(0) = 0. Therefore, we have:
0 = -0^2 + C
C = 0
Substituting C back into the equation, we get:
Δx = -t^2
Finally, we can find the exact change in position of the particle from time t=0 to t=4 seconds by evaluating Δx at t=4:
Δx = -4^2 = -16 cm
Therefore, the exact change in position of the particle from time t=0 to t=4 seconds is -16 centimeters.
if v = 6 - 2t is the velocity
then
s = 6t - t^2 + c, where s is the distance and c is a constant
at t=0 , s = 0-0+c
at t=4 , s = 24 - 16 + c
change in position is
24-16+c - (0+c) = 8 cm