Q.The velocity of a particle moving in a straight line increases uniformly from 5 m/s to 25 m/s in 8 s. Using a velocity-time graph, find

[!TLDR] Acceleration = slope of the v-t line = (25−5)/8; displacement = area of the trapezium under it. [!ANSWER …

Setting up the v-t graph. The velocity increases uniformly (linearly) from $u = 5$ m/s at $t=0$ to $v = 25$ m/s at $t = 8$ s, so the v-t graph is a straight line from the point $(0,5)$ to $(8,25)$.

1. Acceleration. The acceleration is the (constant) slope of this line:

   $a = \frac{v-u}{t} = \frac{25 - 5}{8} = \frac{20}{8} = 2.5\ \text{m/s}^2$

2. Displacement. By Section 2.7, the displacement equals the area under the v-t graph. This region is a trapezium with parallel sides $u = 5$ m/s and $v = 25$ m/s and width (height in the graph's time-axis sense) $t = 8$ s: $s = \frac{1}{2}(u+v)\,t = \frac{1}{2}(5+25)(8) = \frac{1}{2}(30)(8) = 120\ \text{m}$ …

Read acceleration as the slope of the straight-line v-t graph; read displacement as the trapezoidal area under it, and cr …

• Using only the final velocity (25 m/s) to compute displacement instead of the average of initial and final velocity …