Q2-30E
Question
A cat walks in a straight line, which we shall call the \(x\)-axis, with the positive direction to the right. As an observant physicist, you make measurements of this cat’s motion and construct a graph of the feline’s velocity as a function of time (Fig. E2.30). (a) Find the cat’s velocity at \(t = 4.0\,s\) and at \(t = 7.0\,s\).
Step-by-Step Solution
VerifiedHence, the required velocities are
\({v_{4\,s}} = 2.68{\rm{ cm/s and }}{v_{7\,s}} = 1.31{\rm{ cm/s}}\).
The measurement of change of displacement with respect to time is known as velocity, whereas that of velocity with respect to time is known as acceleration.
The given graph for cat's velocity as a function of time is:
From the graph, the acceleration can be evaluated as:
\(a = \frac{{\Delta v}}{{\Delta t}} = \frac{{0 - 8}}{{6 - 0}} \approx - 1.33{\rm{ cm/}}{{\rm{s}}^2}\)
Now, using \({v_t} = {v_0} + at\), at \(t = 4\,s\), the velocity of cat is:
\(\begin{aligned}{v_{4\,s}} = 8 - \left( {1.33 \times 4} \right)\\ = 2.68{\rm{ cm/s}}\end{aligned}\)
Similarly, at \(t = 7\,s\), the velocity of cat is:
\(\begin{aligned}{v_{7\,s}} = 8 - \left( {1.33 \times 7} \right)\\ = 1.31{\rm{ cm/s}}\end{aligned}\)
Hence, the required velocities are
\({v_{4\,s}} = 2.68{\rm{ cm/s and }}{v_{7\,s}} = 1.31{\rm{ cm/s}}\).