Engineering, 24.01.2020 05:31 Olavarriafamily1

Aparticle is placed motionless at a point 12m to the right of the origin. its acceleration as a function of time is a = -30 + 30t. find an expression for the position x of the particle as a function of time t, as it moves along the straight line, also plot the curve x(t). find the total distance travelled from t = 0 to t = 3s.

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Answer from: Kalamazoo373

x(t)=-15t^2+5t^3+12

Explanation:

To find the expression for the position x of the particle as a function of time t you need to integrate the expression for acceleration two times. To find the integral constants you are already given the initial conditions for both the position and the velocity which are:

$x(t=0)=12\\v(t=0)=0$

The first integration gives you the velocity:

$v(t)=\int a(t) dt=-30t+15t^2+C_1$

You get the constant by using the second initial condition:

C_1=0

To get the expression for the position you need to integrate again:

$x(t)=\int v(t)dt=-15t^2+5t^3+C_2$

where $C_2=12\\$ using the first initial condition.

You can now plot the graph. To get the total distance travelled you can integrate the expression for the velocity twice and sum the two integrals (from 0 to 1 and from 1 to 3):

$\int_0^1|v(t)|dt+\int_1^3|v(t)|dt=$ 20 m

Aparticle is placed motionless at a point 12m to the right of the origin. its acceleration as a func

Answer from: acediego3119

Total Distance traveled = 24m

Explanation:

The concept applied in solving this question is the rate of change of motion and its applications in differentiation and integration.

Velocity in terms of rate of change is defined as the rate of change of distance, mathematically, V =ds/dt. this implies that displacement is the integral of velocity with respect to time. Mathematically, S = ∫Vdt.

Acceleration on the other hand is the rate of change of velocity or the second differentiation of the rate of change of displacement. Mathematically, A = dv/dt, as such, velocity is the integral of acceleration with respect to time, i.e V = ∫Adt.

Answer from: Quest

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Answer from: Quest

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