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Standard IX

Physics

Derivation of Velocity-Time Relation by Graphical Method

A car travels...

Question

A car travels a distance of $900$ km with a certain speed. If after travelling half the distance, the speed is increased by $15$ km/h, time required to travel $900$ km is reduced by $1 \frac{1}{2}$ hours. Find the original speed of the car.

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Solution

Let $T_{1} =$ Time taken by car with original speed $= \frac{900}{v}$

$T_{2} =$ Time taken by car when speed is changed after half distance $= \frac{450}{v} + \frac{450}{15 + v}$

Given,
$\Rightarrow T_{1} - T_{2} = \frac{3}{2} \Rightarrow \frac{900}{v} - (\frac{450}{v} + \frac{450}{15 + v}) = \frac{3}{2} \Rightarrow 2 \times 450 \times 15 = 3 \times v \times (v + 15) \Rightarrow v^{2} + 15 v - 4500 = 0 \Rightarrow v = 15, - 30$

Therefore, answer is $15 k m / h r$

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A car travels a distance of 900 km with a certain speed. If after travelling half the distance, the speed is increased by 15 km/h, time required to travel 900 km is reduced by 112 hours. Find the original speed of the car.

Let T1= Time taken by car with original speed=900v T2= Time taken by car when speed is changed after half distance =450v+45015+vGiven, ⇒T1−T2=32⇒900v−(450v+45015+v)=32⇒2×450×15=3×v×(v+15)⇒v2+15v−4500=0⇒v=15,−30Therefore, answer is 15 km/hr

A car travels a distance of $900$ km with a certain speed. If after travelling half the distance, the speed is increased by $15$ km/h, time required to travel $900$ km is reduced by $1 \frac{1}{2}$ hours. Find the original speed of the car.