The velocity of an object as a function of time is given by [tex]v(t) = 2.00 m/s + (3.00 m/s^2)t - (1.0 m/s^3) t ^2[/tex].
Now the instantaneous acceleration of an object with velocity [tex]v(t)[/tex] is the first derivative of the velocity function. That is
[tex]a(t)=\frac{dv(t)}{dt}[/tex]
So,
[tex]a(t)=\frac{d}{dt}[2.00 m/s + (3.00 m/s^2)t - (1.0 m/s^3) t ^2]\\ a(t)=3.00-2t[/tex].
At [tex]t=3[/tex], the instantaneous acceleration is
[tex]a(3)=3.00-2*3=-3 \;\;m/s^2[/tex].
The instantaneous acceleration of the object at time t = 3.00 s is -3.0 m/s²
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]v = 2.00 + 3.00t - 1.0t^2[/tex]
To find instantaneous acceleration, then the velocity needs to be differentiated as follows:
[tex]a = \frac{dv}{dt}[/tex]
[tex]a = \frac{d}{dt} (2.00 + 3.00t - 1.0t^2)[/tex]
[tex]a = 0 + 3.00 - 2.0t[/tex]
[tex]\boxed {a = 3.00 - 2.0t}[/tex]
At t = 3.00 s ,
[tex]a = 3.00 - 2.0t[/tex]
[tex]a = 3.00 - 2.0(3.00)[/tex]
[tex]a = 3.00 - 6.0[/tex]
[tex]\large {\boxed {a = -3.0 ~ m/s^2} }[/tex]
The instantaneous acceleration of the object at time t = 3.00 s is -3.0 m/s²
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate
