3The position of particle at any time t is X=5t-2t2 .find the acceleration of particle at time t=2sec up
Answers
Answer:
The acceleration of the particle at t = 2s is -4 m/s².
Explanation:
Given that,
The x and y coordinates of the particle at any time.
x=5t-2t^2x=5t−2t
2
...(I)
y=10ty=10t ...(II)
Acceleration :
The acceleration is the second derivative of the position of the particle.
Velocity :
The velocity is the firs derivative of the position of the particle.
We need to calculate the velocity of the particle
On differentiating w.r.to t of equation (I)
v_{x}=\dfrac{dx}{dt}=5-4tv
x
=
dt
dx
=5−4t
Again differentiating
a_{x}=\dfrac{d^2x}{dt}=-4a
x
=
dt
d
2
x
=−4
We need to calculate the velocity of the particle
On differentiating w.r.to t of equation (II)
v_{x}=\dfrac{dx}{dt}=10v
x
=
dt
dx
=10
Again differentiating
a_{x}=\dfrac{d^2x}{dt}=0a
x
=
dt
d
2
x
=0
Hence, The acceleration of the particle at t = 2s is -4 m/s².