The particle reach its minimum velocity at time 1.06 sec.
The function is given as
x=5t^3-8t^2+12
Differentiating the above equation with respect to time, to obtain the velocity
dx/dt=v=15t^2-16t
For maximum and minimum values, put dx/dt=0
15t^2-16t=0
On solving the equation, t=0, 1.06
Therefore at time t=1.06 sec, the particle has the minimum value of velocity.
The particle reaches its minimum velocity at t = 0 s or t = 16/15 s
Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration ( m/s² )
v = final velocity ( m/s )
u = initial velocity ( m/s )
t = time taken ( s )
d = distance ( m )
Let us now tackle the problem!
Given:
[tex]x = ( 5t^3 - 8t^2 + 12) ~ m[/tex]
To find the velocity function, we will derive the position function above.
[tex]v = \frac{dx}{dt}[/tex]
[tex]v = 5(3)t^{3-1} - 8(2)t^{2-1}[/tex]
[tex]v = ( 15t^2 - 16t ) ~ m/s[/tex]
Next to calculate the time to reach its minimum speed, then v = 0 m/s
[tex]0 = ( 15t^2 - 16t )[/tex]
[tex]0 = t( 15t - 16)[/tex]
[tex]\large {\boxed {t = 0 ~s ~ or ~ t = 16/15 ~ s} }[/tex]
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle
