A ball is set rolling along level ground with a velocity of 2.8 ms-1. It comes to a halt after travelling a distance of 65 m.
What was the acceleration experienced by the ball during this time?
What direction is the acceleration vector pointing?
What would have caused the acceleration of the ball in this case?
Answers
Answer:
celeration 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
\[x(t)={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2},\]