Question

The acceleration of a particle moving along a straight line is given by a = −kt2 m/s2 where k is a constant and time t is in seconds. The initial velocity of the particle at t = 0 is v0 = 12 m/s and the particle reverses it direction of motion at t = 6 s. Determine the constant k and the displacement of the particle over the same 6-second interval of motion. Ans: k = 1/6 m/s4, Δs = 54 m

Answer 1

Answer:

$X - Xo = 54m$

k = 1/18

Explanation:

Data:

a = -k$t^{2}$$\frac{m}{s^{2} }$

to = 0s    Vo = 12m/s

t = 6s the particle chage it's moviment, so v = 0 m/s

We know that acceleration is the derivative of velocity related to time:

$a = \frac{dV}{dT}$

rearranging...

$a*dT = dV$

Then, we must integrate both sides:

$\int\limits^f_i {dV} \, dV =-k \int\limits^f_i {t^{2} } \, dT$

$V - Vo = -k\frac{t^{3} }{3}$

V = 0 because the exercise says that the car change it's direction:

$0 - 12 = -k\frac{6^{3} }{3}$

k = 1/6

In order to find X - Xo we must integer v*dT = dX

$V - Vo = -k\frac{t^{3} }{3}$

so...

$(Vo -k\frac{t^{3} }{3})dT = dX$

$\int\limits^f_i {dX} \, dX = \int\limits^f_i {Vo -k\frac{t^{3} }{3} } \, dT$

integrating...

$X - Xo = Vot -k\frac{t^{4} }{12}$

$X - Xo = 12*6 -\frac{1}{6}* \frac{6^{4} }{12}$

X - Xo = 54m