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Two particles have positions at time t given by s1 = 4t-t^2 and s2 = 5t^2-t^3. Find the velocities at v1 and v2 at the instant the accelerations of the two particles are equal.

Sagot :

temdan2001

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

  • Instantaneous Velocity
  • Average velocity
  • Relative 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

Learn more about Velocity here: https://brainly.com/question/17228388

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