91Ó°ÊÓ

Acceleration is determined by the rate at which the velocity of an object changes over time.

The quantity of force$\left(F\right)$exerted on an object is equal to the product of its mass $\left(m\right)$and the acceleration generated$\left(a\right)$, according to Newton's second equation of motion.

role="math" localid="1647763785550" $$\begin{gathered} F=ma \\ a=\frac{F}{m} \end{gathered}$$

object's mass, role="math" localid="1647763846467" $\left(m\right)$= $2.0kg$

initial velocityrole="math" localid="1647763903124" $\left(u\right)=0m/s$

force role="math" localid="1647763940636" $\left(F\right)=0N$, at time $t=6s$,

$$\begin{gathered} a=\frac{0}{2} \\ a=0m/s^2 \end{gathered}$$

Hence, acceleration at $t=6s$is$0m/s^2$.

The pace at which an object's location varies in relation to time and a frame of reference.

For an object of mass$\left(m\right)$having initial velocity$\left(u\right)$and final velocity$\left(v\right)$area under force$\left(F\right)$-time$\left(t\right)$graph give change in momentum.

$mv-mu=areaunderforce\left(F\right)-time\left(t\right)graph$

$v=\frac{areaunderforce\left(F\right)-time\left(t\right)graph+mu}{m}$

object's mass, $\left(m\right)$= $2.0kg$

initial velocity $\left(u\right)=0m/s$

force $\left(F\right)=0N$, at time $t=6s$,

$$\begin{gathered} v=\frac{4×3+\left[\left(-2\right)×\left(5-3\right)\right]+2×0}{2} \\ v=\frac{12-4+0}{2} \\ v=\frac{8}{2} \\ v=4m/s \end{gathered}$$

Hence, acceleration at $t=6s$is $4m/s$. These values indicates that the object is moving with uniform velocity.