In this page calculus application problem2 we are going to see solution of the second problem of the topic application problems in calculus.calculus application problem1

**Question 2:**

A particle of unit mass moves so that displacement after t seconds is given by x = 3 cos (2t - 4). Find the acceleration and kinetic energy at the end of 2 seconds. (K.E = (1/2) m v²)

**Solution:**

Here the variable "x" represents the displacement of the object and "t" displacement of a particle in seconds

x (t) = 3 cos (2t - 4)

mass = 1

to find acceleration we have to different the given equation two times

velocity dx/dt = 3 [- sin (2t - 4)] (2(1) - 0)

(v) = - 6 sin (2t - 4)

velocity at t = 2

v = -6 sin (2(2) - 4)

= -6 sin (4-4)

= -6 sin (0)

= 0

acceleration d²x/dt² = -6 cos (2 t - 4) (2(1)- 0)

= -6 cos (2 t - 4) 2

= -12 cos (2 t - 4)

now we have to put t = 2

= -12 cos (2(2) - 4)

= -12 cos (4 - 4)

= -12 cos (0)

= -12 (1)

= -12

Kinetic energy K.E = (1/2) m v²

= (1/2) (1) (0)²

= (1/2) (0)

= 0

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- First Principles
- Implicit Function
- Parametric Function
- Substitution Method
- logarithmic function
- Product Rule
- Chain Rule
- Quotient Rule
- Rolle's theorem
- Lagrange's theorem
- Finding increasing or decreasing interval
- Increasing function
- Decreasing function
- Monotonic function
- Maximum and minimum
- Examples of maximum and minimum