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Marrrta [24]
3 years ago
14

A particle travels along the x-axis in such a way that its acceleration at time t is a(t) = t + t2. if it starts at the origin w

ith an initial velocity of 3 (that is, s(0) = 0 and v(0) = 3), determine its position and velocity when t = 5. (round your answers to two decimal places.)
Physics
1 answer:
Olegator [25]3 years ago
6 0
The acceleration of the particle as a function of time t is
a(t) = t + t^2
The velocity of the particle at time t is the integral of the acceleration:
v(t) =  \int {a(t)} \, dt =  \frac{t^2}{2} +  \frac{t^3}{3}  + C
where the constant C can be found by requiring that the velocity at time t=0 is v=3:
v(0) = 3
and we find C=v_0=3
so the velocity is
v(t)=3+ \frac{t^2}{2}+ \frac{t^3}{3}

The position of the particle at time t is the integral of the velocity:
x(t)=\int {v(t) } \,dt = 3t +  \frac{t^3}{6}+ \frac{t^4}{12}   +D
where D can be found by requiring that the initial position at time t=0 is zero:
x(0)=0
from which we find D=0, so 
 x(t)=3t + \frac{t^3}{6}+ \frac{t^4}{12} 

To solve the problem, now we just have to substitute t=5 into x(t) and v(t) to find the position and the velocity of the particle at t=5.

The position is:
x(5)=3(5) +  \frac{5^3}{6}+ \frac{5^4}{12}=87.92
and the velocity is:
v(5) = 3+ \frac{5^2}{2}+ \frac{5^3}{3}=57.17
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Explanation:

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