Question

A car decelerates uniformly from 20ms to rest in 12seconds, then reverses with uniform acceleration to its original starting point also in 4seconds.
a.Draw a v-t graph

Answer 1

Answer on Question 72732, Physics / Mechanics | Relativity Question

A car decelerates uniformly from 20ms to rest in 12seconds, then reverses with uniform acceleration to its original starting point also in 4seconds.

a. Draw a v-t graph

Solution.

At initial time, $t = 0$ s, the velocity is $v = 12$ m/s. Put the point 1 on the graph that corresponds to these values. In $t = 12$ s the velocity is $v = 0$ m/s. Put the point 2 on the graph that corresponds to these values. Since a car decelerates uniformly draw a line 1-2. The area between the line and the time axis on a velocity-time graph, between two times is the displacement change $\Delta x$ during that time interval. We find that

$$\Delta x = \frac {1}{2} \cdot 2 0 \mathrm {m / s} \cdot 1 2 \mathrm {s} = 1 2 0 \mathrm {m}$$

After stopping the car moves in opposite direction to the same distance in 4 seconds with uniform acceleration to its original starting point. So we can find the velocity at its original starting point from the equation

$$\Delta x = \frac {1}{2} \cdot v \mathrm {m / s} \cdot 4 \mathrm {s} = - 1 2 0 \mathrm {m}$$

The minus sign means that the displacement occurs in the opposite direction. Solving this equation we get $v = -60 \, \text{m/s}$ . Put the point 3 on the graph that corresponds to $t = 12 + 4 = 16 \, \text{s}$ and $v = -60 \, \text{m/s}$ . Then draw a line 2-3.

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