Answer:
Explanation:
Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocity function is acceleration,
\[\frac{d}{dt}v(t)=a(t),\]
we can take the indefinite integral of both sides, finding
\[\int \frac{d}{dt}v(t)dt=\int a(t)dt+{C}_{1},\]
where C1 is a constant of integration. Since
\[\int \frac{d}{dt}v(t)dt=v(t)\]
, the velocity is given by
\[v(t)=\int a(t)dt+{C}_{1}.\]
Similarly, the time derivative of the position function is the velocity function,
\[\frac{d}{dt}x(t)=v(t).\]
Thus, we can use the same mathematical manipulations we just used and find
\[x(t)=\int v(t)dt+{C}_{2},\]
where C2 is a second constant of integration.
We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doing the integration in (Figure), we find
\[v(t)=\int adt+{C}_{1}=at+{C}_{1}.\]
If the initial velocity is v(0) = v0, then
\[{v}_{0}=0+{C}_{1}.\]
Then, C1 = v0 and
\[v(t)={v}_{0}+at,\]
which is (Equation). Substituting this expression into (Figure) gives
\[x(t)=\int ({v}_{0}+at)dt+{C}_{2}.\]
Doing the integration, we find
\[x(t)={v}_{0}t+\frac{1}{2}a{t}^{2}+{C}_{2}.\]
If x(0) = x0, we have
\[{x}_{0}=0+0+{C}_{2};\]
so, C2 = x0. Substituting back into the equation for x(t), we finally have
\[x(t)={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2},\]
Answer:
B. The object has negative acceleration
Explanation:
A. The object cannot have negative displacement as it is moving.
C. The object has a displacement as it is moving.
D. The object is accelerating because of movement.
The correct option is B. The object has negative acceleration as it is moving and it's speed must be decreasing or it may be stopping. The other choices are not applicable.
