Is it ever possible that the object is moving with a velocity such that its rate of change of speed is not constant, but rate of change of velocity is constant?
Like speed is only the magnitude, so how can its change be non constant, but rate of change of velocity be constant?
Hint: In the projectile motion (without drag) the acceleration $\vec{a}=\frac{d\vec{v}}{dt}$ is constant. However $\frac{d|\vec{v}|}{dt}$ is not constant, since it is negative when the projectile is rising and positive when the projectile is falling.
A nice fact which may be a bit counterintuitive, is that if you have a square with diagonal length $\ell$ and this length varies in time, the second time derivative of the area of that square can be constant, while the second time derivative of the diagonal need not be constant.
The simplest example we can use to illustrate this point, is probably:
$$ \ell(t)=\sqrt{1+t^2} \tag{1} $$
The area of a square is related to its diagonal by $A=\frac{1}{2}\ell^2$. So, first let's verify that the area given by $A(t)=\frac{1}{2}\ell^2(t)$ has a constant second time derivative:
$$ \dot{A}(t)=\frac{\mathrm d}{\mathrm dt}\left(\frac{1}{2}(1+t^2)\right)=t \tag{2}$$ $$ \ddot{A} = 1 $$
On the other hand, we see that:
$$ \dot{\ell}(t) = \frac{t}{\sqrt{1+t^2}}, $$ \begin{align*} \ddot{\ell}(t) &= \frac{\sqrt{1+t^2}-\large\frac{t^2}{\sqrt{1+t^2}}}{1+t^2} \\[6pt]&= \frac{1}{\Big(1+t^2\Big)^{\large\frac{3}{2}}} \tag{3} \end{align*}
Which is manifestly not constant in time.
I claim that this has a direct relevance to your question, and that it's also directly related to the answer given by @Qmechanic here.
If we take the velocity vector corresponding to a projectile moving under the influence of gravity, we can generally express it as:
$$ \vec{v} = v_{_{\large 0x}}\hat{x}+(v_{_{\large 0y}}-gt)\hat{y} \tag{4} $$
Note that its magnitude is given by:
$$ |\vec{v}| = \sqrt{v_{_{\large 0x}}^2+v_{_{\large 0y}}^2+g^2t^2-2v_{_{\large 0y}}gt} \ , $$
but this expression is a bit too messy in order to illustrate the conceptual point. So let's choose $v_{_{\large 0y}}=0$, corresponding to the case where the object is projected from some height above ground so that its initial velocity is purely horizontal. Then $|\vec{v}|$ becomes:
$$ |\vec{v}| = \sqrt{v_{_{\large 0x}}^2+g^2t^2} \tag{5} $$
Which has exactly the same form, up to a choice of units and an overall multiplicative factor, as $(1)$ above. Therefore this velocity vector indeed satisfies both conditions you mention:
$$\frac{\mathrm d^2 |\vec{v}|^2 }{\mathrm d t^2}=\frac{\mathrm d^2 }{\mathrm d t^2}\left(v_{_{\large 0x}}^2+g^2t^2\right)=2g^2\ , \tag{6}$$
But wait, am I comparing apples and oranges? What is the relation between $\large\frac{\mathrm d^2 |\vec{v}|^2 }{\mathrm d t^2}$ being constant and $\dot{\vec{v}}$ being constant? Well, the relation can be shown by considering the time derivative of the dot product of $\vec{v}$ with itself:
$$ \frac{\mathrm d}{\mathrm dt}(\vec{v}\cdot\vec{v}) = \frac{\mathrm d }{\mathrm d t}(|\vec{v}|^2) = 2\dot{\vec{v}}\cdot\vec{v} $$
$$ \Rightarrow \frac{\mathrm d^2 }{\mathrm d t^2}(|\vec{v}|^2) = 2\Big|{\dot{\vec{v}}}\Big|^2+2\ddot{\vec{v}}\cdot\vec{v} \tag{7}$$
From this you immediately see that since $\dot{\vec{v}}$ is constant, we have $\ddot{\vec{v}}=\vec{0}$ and therefore the left hand side, which is the second time derivative of the magnitude squared again (like in our example with the square), is constant. You can immediately see how that confirms our calculation in $(6)$ by putting $\dot{\vec{v}}=-g\ \hat{y}$ in this last equation.
Overall, we have shown that the fact that a vector can have a constant time derivative, while its magnitude has a non-constant time derivative, is geometrically analogous to a square that has a varying diagonal while the second time derivative of its area is constant.
In fact, the velocity vector of a projectile behaves in exact analogy to the geometrical example we started with. The area it describes, which corresponds to the square of its magnitude, has a constant second time derivative, and we have shown by $(7)$ that this is a direct consequence of $\dot{\vec{v}}=\text{constant}$; its magnitude $|\vec{v}|$ however, has a non-constant time derivative, which isn't very surprising given the effect of the gravitational acceleration on it.
In fact we can see that for a projectile, $\large\frac{\mathrm d|\vec{v}|}{\mathrm dt}$ is constant only under the condition $v_{_{\large{0x}}}=v_{_{\large{0y}}}=0$, corresponding to free fall and an absence of any initial velocity.
It is important to notice a subtlety in equation $(7)$, which is that the converse of what we've shown is not necessarily true: if $\large\frac{\mathrm d^2 |\vec{v}|^2 }{\mathrm d t^2}$ is constant, the quantity on the right hand side is a sum of two contributions, only one of which involves $\dot{\vec{v}}$. Therefore we see that:
$$\dot{\vec{v}} = \text{constant} \color{blue}{\boldsymbol{\Rightarrow}} \frac{\mathrm d^2 |\vec{v}|^2 }{\mathrm d t^2}= \text{constant}, \ $$ however, $$ \frac{\mathrm d^2 |\vec{v}|^2 }{\mathrm d t^2}= \text{constant}\color{red}{\boldsymbol{\nRightarrow}}\dot{\vec{v}} = \text{constant} \ (!) $$
This doesn't subtract anything from the validity of neither of our examples (the square one nor the projectile one), because in both we in fact have a vector that has a constant time derivative (can you see why that's true for the case of the square that varies in time, too? ;).
This only serves to caution the reader not to assume the two conditions are equivalent.
Yes, this happens all the time. Fire a gun, or throw a ball, or do just about anything that involves making something move. And ignore things like air resistance, curvature of the earth and so on.
As soon as the bullet leaves the barrel, it's under the influence of constant gravitational acceleration. But initially, assuming you fired horizontally, that acceleration is perpendicular to the bullet's motion, so it doesn't change the bullet's speed. After a while, as the bullet starts to descend towards the ground, that same acceleration increases the bullet's speed.
In general if $v$ denotes the velocity, the rate of change of speed is
\begin{align*}\frac{\text{d}|v|}{\text{d}t} &= \frac{\text{d}}{\text{d}t} \sqrt{ \left< v, v \right> } \\&= \frac{1}{2 \sqrt{ \left< v, v \right> }} \cdot \frac{\text{d}}{\text{d}t} \left< v, v \right> \\&= \frac{\left< \dot{v}, v \right>}{|v|} \\&= |\dot{v}| \cdot \cos \angle( \dot{v}, v).\end{align*}
We can see that the rate of change of speed ($\frac{\text{d}|v|}{\text{d}t}$) and [the magnitude of] the rate of change of velocity ($|\dot{v}|$) differ by a factor of $\cos \angle( \dot{v}, v)$. If we apply a constant acceleration that is not perfectly aligned with velocity, that angle will change over time - their directions will gradually come together, just like the velocity of a free falling object gradually becomes vertical. It then follows from the formula that the rate of change of speed will not be constant.
