Question #217324
A car runs at a constant speed of 15m/s for 300s and then accelerate uniformly to a speed of 25m/s over a period of 20s. This is maintained for 300s before the car is brought to rest with uniform deceleration in 30s. Draw a velocity time graph to represent the journey described above from the graph find :(i) the acceleration while velocity changes from 15m/s to 25/s.(ii) the total distance traveled in the time described. (iii) the average speed over the time described"
1
Expert's answer
2021-07-19T09:49:25-0400

Explanations & Calculations




  • Acceleration is given by the slope. Then,

a=2515320300=1020=0.5ms2\qquad\qquad \begin{aligned} \small a&=\small \frac{25-15}{320-300}=\frac{10}{20}=\bold{0.5\, ms^{-2}} \end{aligned}


  • Distance is given by the area under the graph for a given period of time. Then the total distance is,

A=A1+A2+A3+A4=(15×300)+(0.5(15+25)(320300))+(25×(620320))+(0.5(25)(650620))=12775m\qquad\qquad \begin{aligned} \small A&=\small A_1+A_2+A_3+A_4\\ &=\small (15\times300)+(0.5(15+25)(320-300))+(25\times(620-320))+(0.5(25)(650-620))\\ &=\small \bold{12775\,m} \end{aligned}


  • Average speed is the total distance by total time. then,

vavg=12775m650s=19.65m\qquad\qquad \begin{aligned} \small v_{avg}&=\small \frac{12775\,m}{650\,s}\\ &=\small \bold{19.65m} \end{aligned}


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