Answer to Question #261379 in Molecular Physics | Thermodynamics for Noob

Question #261379

A bus starts from rest at a station and accelerates at a speed rate of 2m/s^2 for 15 seconds. It then runs at a constant speed for 25 seconds and slows down at -3m/s^2 until it stops. Find the total distance travelled by the bus.

1
Expert's answer
2021-11-07T19:29:27-0500

Let's first find the displacement of the bus when it accelerates:


d1=v0t+12at2,d_1=v_0t+\dfrac{1}{2}at^2,d1=0+12×2 ms2×(15 s)2=225 m.d_1=0+\dfrac{1}{2}\times2\ \dfrac{m}{s^2}\times(15\ s)^2=225\ m.

Then, we can find the displacement of the bus when it runs at a constant speed:


v=v0+at=0+2 ms2×25 s=50 ms,v=v_0+at=0+2\ \dfrac{m}{s^2}\times25\ s=50\ \dfrac{m}{s},d2=v0t=50 ms×25 s=1250 m.d_2=v_0t=50\ \dfrac{m}{s}\times25\ s=1250\ m.

Let's find the time that the bus takes to stop:


v=v0+at,v=v_0+at,t=vv0a=0 ms50 ms3 ms2=16.7 s.t=\dfrac{v-v_0}{a}=\dfrac{0\ \dfrac{m}{s}-50\ \dfrac{m}{s}}{-3\ \dfrac{m}{s^2}}=16.7\ s.

Then, we can find the displacement of the bus when it decelerates:


d3=v0t+12at2,d_3=v_0t+\dfrac{1}{2}at^2,d3=50 ms×16.7 s+12×(3 ms2)×(16.7 s)2=417 m.d_3=50\ \dfrac{m}{s}\times16.7\ s+\dfrac{1}{2}\times(-3\ \dfrac{m}{s^2})\times(16.7\ s)^2=417\ m.

Finally, we can find the total displacement travelled by the bus:


dtot=d1+d2+d3=225 m+1250 m+417 m=1892 m.d_{tot}=d_1+d_2+d_3=225\ m+1250\ m+417\ m=1892\ m.

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