Answer :
- first trip: the distance is
S=90 m
while the speed is
v=56 m/h
Therefore the time taken is
[tex]t_1= \frac{S}{v_1}= \frac{90 m}{56 m/h}= 1.6 h[/tex]
- second trip: the distance covered is still the same,
S=90 m
while the average speed this time was
v=49 m/h
Therefore, the time taken for the second trip was
[tex]t_2 = \frac{S}{v_2}= \frac{90 m}{49 m/h}=1.8 h [/tex]
- The average speed for the whole trip is the total distance divided by the total time taken:
[tex]v= \frac{S_1 + S_2 }{t_1+t_2}= \frac{90 m+90 m}{1.6 h+1.8 h}=52.9 m/h [/tex]
S=90 m
while the speed is
v=56 m/h
Therefore the time taken is
[tex]t_1= \frac{S}{v_1}= \frac{90 m}{56 m/h}= 1.6 h[/tex]
- second trip: the distance covered is still the same,
S=90 m
while the average speed this time was
v=49 m/h
Therefore, the time taken for the second trip was
[tex]t_2 = \frac{S}{v_2}= \frac{90 m}{49 m/h}=1.8 h [/tex]
- The average speed for the whole trip is the total distance divided by the total time taken:
[tex]v= \frac{S_1 + S_2 }{t_1+t_2}= \frac{90 m+90 m}{1.6 h+1.8 h}=52.9 m/h [/tex]