843 wrote:
I have a problem with displacement. I often can't figure out how to get the total distance from displacement.
A particle, travelling in a straight line, passes a fixed point O on the line with a speed of 0.5 m/s. The acceleration, a m/s, of the particle, t s after passing O, is given by a = 1.4 - 0.6t.
Find the total distance travelled by the particle between t=0 and t=10
We know that acceleration is the time derivative of velocity and displacement is the time derivative of velocity. The difference between displacement and distance, is that distance is a scalar; in one-dimension displacement is the velocity * time; distance is the absolute value of velocity (speed) * time.
a(t) = 1.4 -0.6t
v(t) = ∫a(t)dt = ∫(1.4 - 0.6t)dt = 1.4t - 0.3t^2 + C
Solve for C by using your knowledge that v(0) = 0.5
v(0) = 1.4*0 - 0.3*0^2 + C
0.5 = C
Thus, v(t) = 1.4t - 0.3t^2 + 0.5
Distance, we'll call it f(t) = ∫|v(t)dt| = ∫|1.4t - 0.3t^2 + 0.5 dt|. The limits of integration here are 0 and 10 (as you are concerned about the problem from t = 0 to t = 10).
Integrating an absolute value function is probably easiest by splitting it up into multiple integrals.
∫|0.3t^2 + 1.4t + 0.5 dt| =
∫0.3t^2 + 1.4t + 0.5 dt ]t=0 to t=5 + ∫-(0.3t^2 + 1.4t + 0.5)dt ]t=5 to t=10.
If the splitting or the notation is confusing, let me know.
Anyway, now it's just a matter of integrating.
[0.1t^3 + 0.7t^2 + 0.5t] t=0 to t=5 - [0.1t^3 + 0.7t^2 + 0.5t] t=5 to t=10.
0.1*5^3 + 0.7*5^2 + 0.5*5 - (0.1*5^3 + 0.7*5^2 + 0.5*5 - (0.1*10^3 + 0.7*10^2 + 0.5*10)) = 175 m