Physics - Kinematics Question with Solution | TestHub

$2t=\sqrt{t+3+x}$.

$4t^2=t+3+x$

$x=4t^2-t-3$

$v=\frac{dx}{dt}=8t-1$

$a=\frac{dv}{dt}=8$ → (B)

At $t=1$:

$x=4(1{)}^2-1-3=0$ → (C)

$\Delta x=x\left(\frac{1}{4}\right)-x(0)=-3-(-3)=0$ →(D)

Velocity is not constant, as $v=8t-1$ depends on $t$. Therefore, (A) is false.