Finding distance by integrating acceleration function.?

Question

Hi everyone,

Bit stuck on the following issue and have included working out and answer in brackets but stuck as to how my tutor concluded with such an answer.

"A body of mass 100g moves in a straight line with initial velocity as 10 m/s. The deceleration of the body is given by a = a0 - kt; where a0 = 0.8m/s^2 and k = 0.4m/s^3"

a) find the distance the body will travel in 2 seconds

I know how to find distance, you need to integrate acceleration once to get velocity then once again to find distance. Here are the answers of such with my queries in brackets.

V(t) = a0t - 0.5kt^2 + C (Where does the 0.5 come from)

S(t) = 0.5a0 t^2 - 1/6 kt^3 +10t (integrating velocity above does not give 1/6? and why is 0.5 at the start of the formula now, is it rearranged for a reason)

Thanks so much.

Answer 1

In order to get velocity you have, as you correctly state, to integrate acceleration.

acceleration equals ao - kt. When you integrate it (w.r to t) you get a0t -k(t^2)/2 + Vo.

Recall, 0.5 and 1/2 are the same quantity and this is where 0.5 comes from.

So, velocity V(t) = Vo + a0t - -k(t^2)/2

Second step, as you correctly state, is to integrate V(t) :

S(t) = S(0) + Vot + a0(t^2)/2 - (k/2)(t^3)/3 = S(0) + Vot + a0(t^2)/2 - k/(t^3)/6

And the Answer to your problem is : V(2) - V(0) = 2Vo + 2a0 - 4k/3.

What is left for you is to plug in the numerical values of Vo, a0 & k.

Answer 2

dv/dt = a = a0 - kt

Integrate remembering integral of t is (1/2)t^2 + constant

dS/dt = v(t) = a0t - 0.5kt^2 + c

The 0.5 comes from the fact that integral of k*t is k*(1/2)t^2 = 0.5kt^2

Initial velocity 10 m/s means when t = 0, v = 10

c = 10

dS/dt = a0t - (1/2)kt^2 + 10

Integrate remembering integral of t^2 is (1/3)t^3 + other constant

S = (1/2)a0 t^2 - (1/6)kt^3 + 10t + C



Note: The reason that 0.5 is at the beginning of the formula is that integral of t is 0.5t^2



a0 = 0.8m/s^2, so, a0/2 = 2/5

k = 0.4m/s^3,so, k/6 = 1/15

S = (2/5)t^2 - (1/15)t^3 + C



a) When t = 0, S = C, so neglect C in calculation of distance in 2 secs.

S = (2/5)2^2 - (1/15)2^3 = 16/15 metres



Regards - Ian

Answer 3

It looks like you've just forgotten the power rule for integration and need to review. Remember, the derivative of t^n is n t^(n-1), and the integral of t^n is t^(n+1)/(n+1)



So the 0.5's come from integrating t to get t^2/2, and the 1/6 comes from integrating (1/2)t^2 to get (1/2)(t^3/3)