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# A train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/h from its usual speed. Find the usual speed of the train.

Last updated date: 10th Sep 2024
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Hint: Think of the basic definition of speed and focus on the point that the product of speed and time is equal to the distance covered and the distance covered for the cases mentioned in the above question is 300km. Start by taking the usual speed and time to be v km/hr and t hours.

Let’s start with what is speed. Speed is a scalar quantity defined as the distance travelled by a particle or object per unit time.
Generally, we deal with two kinds of speeds. One is instantaneous, and the other is the average speed. For uniform motion, both are identical.
Average speed is defined as the total distance covered by a body divided by the time taken by the body to cover it.
$\therefore {{v}_{avg}}=\dfrac{\text{distance covered}}{\text{time taken}}$
Now, starting with the solution to the above question. Let the actual time taken be t hours and the distance to be travelled be x km.
When the train was moving with a speed of v km/hour, it takes t hours to cover a distance of 300km. We can represent this in the form of mathematical equation as:
$t=\dfrac{\text{distance covered}}{\text{speed}}=\dfrac{300}{v}..........(i)$
Also, it is mentioned that the train takes 2 hours less when the speed is increased by 5 km/hr.
$t-2=\dfrac{\text{distance covered}}{\text{speed}}=\dfrac{300}{v+5}.........(ii)$
Now we will subtract equation (ii) from equation (i). On doing so, we get
$t-\left( t-2 \right)=\dfrac{300}{v}-\dfrac{300}{v+5}$
$\Rightarrow 2=\dfrac{300\left( v+5 \right)-300v}{v\left( v+5 \right)}$
$\Rightarrow 2v\left( v+5 \right)=1500$
$\Rightarrow {{v}^{2}}+5v=750$
$\Rightarrow {{v}^{2}}+5v-750=0$
Now, we know that 5v=30v-25v. So, if we use this in our equation, we get
${{v}^{2}}+30v-25v-750=0$
$\Rightarrow v\left( v+30 \right)-25\left( v+30 \right)=0$
$\Rightarrow \left( v+30 \right)\left( v-25 \right)=0$
So, the possible values of v according to the quadratic equation are: -30 km/hr and 25 km/hr. We know that speed cannot be negative. So, the acceptable value of v is 25km/hr.
So, we can conclude that the actual speed of the train is 25km/hr.

Note: Make sure that you convert all the elements required for solving the problem to a standard unit system to avoid errors. Also, remember that speed cannot be negative but velocity can be, so be careful about what is mentioned in the question.