The problem asks us to find the time taken for a car to travel a certain distance at a new speed, given the initial speed and time taken.
The core relationship is between distance, speed, and time: Distance = Speed × Time.
If the distance remains constant, speed and time are inversely proportional. This means if the speed decreases, the time taken increases proportionally.
Calculate the Total Distance: First, find the distance of the journey using the initial speed and time.
Initial Speed ($v_1$) = $50 \text{ km/hour}$
Initial Time ($t_1$) = $8 \text{ hours}$
Distance ($d$) = $v_1 \times t_1 = 50 \text{ km/h} \times 8 \text{ h} = 400 \text{ km}$.
Calculate the New Time: Now, use the same distance and the new speed to find the time required.
New Speed ($v_2$) = $40 \text{ km/hour}$
Distance ($d$) = $400 \text{ km}$
New Time ($t_2$) = $\frac{d}{v_2} = \frac{400 \text{ km}}{40 \text{ km/h}} = 10 \text{ h}$.
Alternatively, using the inverse proportion:
$v_1 \times t_1 = v_2 \times t_2$
$50 \text{ km/h} \times 8 \text{ h} = 40 \text{ km/h} \times t_2$
$400 = 40 \times t_2$
$t_2 = \frac{400}{40} = 10 \text{ h}$.
It will take 10 hours to travel the same distance at a speed of 40 km/hour.
A journey of 900 km is completed in 11 h. If two-fifth of the journey is completed at the speed of 60 km/h, at what speed (in km/h) is the remaining journey completed?
A car starts from point A towards point B, travelling at the speed of 20 km/h. 1 \(\frac{1}{2}\) hours later, another car starts from point A and travelling at the speed of 30 km/h and reaches 2 \(\frac{1}{2}\) hours before the first car. Find the distance between A and B.
A bus covered a distance of 162 km. If speed of this bus is 15 m/s, then what will be the time taken ?
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