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Q.Show that the velocity of uniform motion of an object is equal to the slope of position-time graph.

Nagaland NbseNagaland Board of School Education (Class XI) 2022Subjective· 2mImportance★★★★★
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The slope of a straight-line position–time graph equals the object's constant velocity.

For an object in uniform motion, velocity vv is constant, so its position at time tt is

x(t)=x0+vtx(t) = x_0 + vt

where x0x_0 is the position at t=0t=0. This is the equation of a straight line when xx is plotted against tt, with x0x_0 as the intercept and vv as the coefficient of tt.

Consider two instants t1t_1 and t2t_2, with corresponding positions x1=x0+vt1x_1 = x_0+vt_1 and x2=x0+vt2x_2 = x_0+vt_2. The slope of the chord (and, since the graph is a straight line, of the graph everywhere) between these points is

slope=x2−x1t2−t1=(x0+vt2)−(x0+vt1)t2−t1=v(t2−t1)t2−t1=v\text{slope} = \frac{x_2-x_1}{t_2-t_1} = \frac{(x_0+vt_2)-(x_0+vt_1)}{t_2-t_1} = \frac{v(t_2-t_1)}{t_2-t_1} = v

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