Question #266177

1. Draw the velocity-time graph of a car that starts with an initial velocity of 15km/hr and accelerates uniformly at 5ms-2 until it attains a maximum velocity of 50km/hr, it them maintains this speed for the next 2min. Use the graph to determine:



a. The total distance covered before it maintains the maximum velocity reached



b. The total distance covered in the entire journey



c. Average velocity

1
Expert's answer
2021-11-15T10:33:26-0500

Let's first find the time that the car takes to attains a maximum velocity of 13.89 m/s:


v=v0+at,v=v_0+at,t=vv0a,t=\dfrac{v-v_0}{a},t=(50 kmh15 kmh)×1000 m1 km×1 h3600 s5 ms2=1.94 s.t=\dfrac{(50\ \dfrac{km}{h}-15\ \dfrac{km}{h})\times\dfrac{1000\ m}{1\ km}\times\dfrac{1\ h}{3600\ s}}{5\ \dfrac{m}{s^2}}=1.94\ s.

Let's draw the velocity versus time graph:



(a) We can find total distance covered before it maintains the maximum velocity reached from the area of small triangle under the graph:


d1=12bh=12×1.94 s×(13.89 ms4.17 ms)=9.43 m.d_1=\dfrac{1}{2}bh=\dfrac{1}{2}\times1.94\ s\times(13.89\ \dfrac{m}{s}-4.17\ \dfrac{m}{s})=9.43\ m.

(b) Let's first find the distance covered by the car for the next 2 min. It can be found from the area of rectangle under the graph:


d2=ab=13.89 ms×(121.94 s1.94 s)=1667 m.d_2=ab=13.89\ \dfrac{m}{s}\times(121.94\ s-1.94\ s)=1667\ m.

Finally, we can find the total distance covered in the entire journey:


dtot=d1+d2=9.43 m+1667 m=1676.43 m.d_{tot}=d_1+d_2=9.43\ m+1667\ m=1676.43\ m.

(c) We can find the average velocity as follows:


vavg=dtotttot=1676.43 m121.94 s=13.75 ms.v_{avg}=\dfrac{d_{tot}}{t_{tot}}=\dfrac{1676.43\ m}{121.94\ s}=13.75\ \dfrac{m}{s}.

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