A bus moves from rest with a uniform acceleration of 2 m/s2 for the first 10 s. It then accelerate at a uniform rate of 1 m/s2 for another 15 s. It continues at constant speed for 70 s and finally comes to rest in 20 s by uniform deceleration. Draw its velocity time graph and calculate total distance covered.
Solution
Graph can be drawn as
Are of v-t curve will give displacement
So
Total are of OABCD IS
"S=\\frac{1}{2}\\times20\\times10+((20\\times15)+\\frac{1}{2}\\times15\\times15) \\\\+35\\times70+\\frac{1}{2}\\times20\\times35"
"S=3312.5m"
So distance travelled by bus is 3312.5m
Comments
Thanks for this.
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