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Question: A car covers the first \(2km\) of the total distance of \(4km\) with a speed of \(30km/h\) and the r...

A car covers the first 2km2km of the total distance of 4km4km with a speed of 30km/h30km/h and the remaining distance of 2km2km with a speed of 20km/h20km/h. The average speed of the car is:
A. 24km/h24km/h
B. 25km/h25km/h
C. 50km/h50km/h
D. 26km/h26km/h

Explanation

Solution

Distance covered by anybody is defined as the length of path covered by the person in moving from one point from another. And speed is defined as the rate of change of distance with respect to time.

Complete step by step answer:
Distance: Distance(S)\left( S \right) covered by anybody is defined as the length of path covered by the person in moving from one point to another. Its SI unit is metre(m)\left( m \right).
Speed: Speed(v)\left( v \right) is defined as the rate of change of distance with respect to time. Which means,
v=ΔSΔtv = \dfrac{{\Delta S}}{{\Delta t}}

Instantaneous Velocity: Instantaneous velocity(vinst)\left( {{v_{inst}}} \right) is defined as the derivative of displacement over time at a particular moment in time. In a graph, it is defined as the slope of the graph at that particular time.
Average speed: Average speed(v)\left( {\overline v } \right) is defined as the ratio of total distance covered to total time taken to cover that distance.
So, this case is solved in two parts since there are two speeds given.

First part: Let the distance covered in the first part be S1{S_1}, which is covered in time t1{t_1} with speed v1{v_1}. So,
S1=2km{S_1} = 2km
v1=30km/h\Rightarrow{v_1} = 30km/h
v1=S1t1\because {v_1} = \dfrac{{{S_1}}}{{{t_1}}}
t1=S1v1\Rightarrow {t_1} = \dfrac{{{S_1}}}{{{v_1}}}
t1=230h=115h\Rightarrow {t_1} = \dfrac{2}{{30}}h = \dfrac{1}{{15}}h

Second part: Let the distance covered in the first part be S2{S_2}, which is covered in time t2{t_2} with speed v2{v_2}. So,
S2=2km{S_2} = 2km
v2=20km/h{v_2} = 20km/h
v2=S2t2\because {v_2} = \dfrac{{{S_2}}}{{{t_2}}}
t2=S2v2\Rightarrow {t_2} = \dfrac{{{S_2}}}{{{v_2}}}
t2=220h=110h\Rightarrow {t_2} = \dfrac{2}{{20}}h = \dfrac{1}{{10}}h
So, average speed is,
v=S1+S2t1+t2\overline v = \dfrac{{{S_1} + {S_2}}}{{{t_1} + {t_2}}}
v=4115+110\Rightarrow \overline v = \dfrac{4}{{\dfrac{1}{{15}} + \dfrac{1}{{10}}}}
v=42+330\Rightarrow \overline v = \dfrac{4}{{\dfrac{{2 + 3}}{{30}}}}
v=4×305\Rightarrow \overline v = \dfrac{{4 \times 30}}{5}
v=24km/h\therefore \overline v = 24km/h

Thus the correct answer is option A.

Note: For special cases like this there are simpler formulae to calculate average speed. For a case in which the journey is covered in equal time intervals with different speeds, the average speed can be given by, v=v1+v22\overline v = \dfrac{{{v_1} + {v_2}}}{2}. And when the journey is covered in equal distance intervals with different speeds, the average speed can be given by, v=v1v2v1+v2\overline v = \dfrac{{{v_1}{v_2}}}{{{v_1} + {v_2}}}.