3.27 The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s (b) t = 2 s to 6 s.
For the motion from 5 sec to 10 sec, u = 12 m/s and a = -2.4 m/s2
and t = 5 sec to t = 6 sec means n = 1 for this motion
Distance covered in the 6th sec is S2 = u + (1/2) a (2n-1) = 12 + (1/2) (-2.4) (2 * 1-1) = 10.8 m
Therefore the total distance covered from t = 2 to 6 s = 25.2 + 10.8 m = 36 m
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<p><strong>3.27 </strong></p><p><strong> (a)</strong> Distance travelled by the particle between t = 0 s and t = 10 s is the area of the triangle = (1/2) x base x height = (1/2) x 10 x 12 = 60 m</p><p>The average speed of the particle is 60/10 m/s = 6 m/s</p><p> </p><p><strong> (b)</strong> Distance travelled by the particle between t = 2 s and t = 6 s</p><p>Let S<sub>1</sub> be the distance travelled by the particle in time 2 to 5 s and S<sub>2</sub> be the distance travelled between 5 to 6s.</p><p>For the motion from 0 to 5 sec, u = 0, t = 5, v = 12 m/s</p><p>From the equation v = u + at we get a = (v-u)/t = 12/5 = 2.4 m/s<sup>2</sup></p><p>Distance covered from 2 to 5 sec, S<sub>1</sub> = distance covered in 5 s – distance covered in 2 s</p><p>From the equation s = ut + at<sup>2</sup> we get, </p><p>= (1/2) * 2.4 * 25 – (1/2) * 2.4 * 4 = 30 – 4.8 = 25.2 m</p><p>For the motion from 5 sec to 10 sec, u = 12 m/s and a = -2.4 m/s<sup>2</sup></p><p>and t = 5 sec to t = 6 sec means n = 1 for this motion</p><p>Distance covered in the 6th sec is S2 = u + (1/2) a (2n-1) = 12 + (1/2) (-2.4) (2 * 1-1) = 10.8 m</p><p>Therefore the total distance covered from t = 2 to 6 s = 25.2 + 10.8 m = 36 m</p>
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