The Speed of onward and return journey will be "20 mph and 10 mph".
Let,
According to the question,
Time taken,
- t₁ = [tex]\frac{40}{S+10}[/tex]
Return Journey travel time,
- t₂ = [tex]\frac{40}{S}[/tex]
then,
→ [tex]\frac{40}{S} -\frac{40}{S+10} =2[/tex]
S = 10, -20 (negative)
→ [tex]t_1 = \frac{40}{20}[/tex]
[tex]= 2 \ hr[/tex]
→ [tex]t_2 = \frac{40}{10}[/tex]
[tex]= 4 \ hr[/tex]
hence,
→ The speed for onward journey will be:
= [tex]\frac{40}{2}[/tex]
= [tex]20 \ mph[/tex]
→ The speed for return journey will be:
= [tex]\frac{40}{4}[/tex]
= [tex]10 \ mph[/tex]
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