By differentiating the two equations, the velocities at v1 and v2 at the instant the accelerations of the two particles are equal are 0 m/s and 8 m/s
VELOCITY
There are various type of velocity
Given that two particles have positions at time t given by s1 = 4t-t^2 and
s2 = 5t^2-t^3.
The velocities at v1 and v2 at the instant the accelerations of the two particles are equal can be calculated by first differentiating the two equations
s1 = 4t-t^2
[tex]\frac{ds}{dt}[/tex] = 4 -2t
Acceleration will be second derivative
[tex]\frac{dv}{dt}[/tex] = -2
for the second equation: s2 = 5t^2-t^3
[tex]\frac{ds}{dt}[/tex] = 10t - 3[tex]t^{2}[/tex]
[tex]\frac{dv}{dt}[/tex] = 10 - 6t
Since the acceleration are equal, equate the two together
-2 = 10 - 6t
collect the likes term
-2 - 10 = -6t
-12 = -6t
t = 12/6
t = 2 s
Substitute t in the first ds/dt equation
[tex]V_{1}[/tex] = 4 - 2t
[tex]V_{1}[/tex] = 4 - 2(2)
[tex]V_{1}[/tex] = 0
Substitute t in the second ds/dt equation
[tex]V_{2}[/tex] = 10t - 3[tex]t^{2}[/tex]
[tex]V_{2}[/tex] = 10(2) - 3(2)^2
[tex]V_{2}[/tex] = 20 - 12
[tex]V_{2}[/tex] = 8 m/s
Therefore, the velocities at v1 and v2 at the instant the accelerations of the two particles are equal are 0 m/s and 8 m/s
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