Speed/Distance

vaibhav101 wrote:A motorcyclist rode the first half of his way at a constant speed. Then he was delayed for 5 minutes and, therefore, to make for the lost time he increased his speed by 10 km/hr. Find the initial speed of the motorcyclist if the distance covered by him is equal to 50 km.

A 36 km/hr
B 48 km/hr
C 50 km/hr
D 62 km/hr
E 55 km/hr

Hello vaibhav101.

This is the way I'd solve this PS question.

We have that:

d = 50km.
v_1 - initial velocity
v_2 = v_1+10km/h
t_1 - time first 25km
t_2 = t_1-5 minutes

Since we have to calculate km/h we have to convert 5 minutes to hours. Then 5 minutes is the same as 1/12 hours. Therefore,

t_2=t_1 - 1/12.

Also, we know that v_1*t_1=25 km.

Therefore, we have $$d=v\cdot t\ $$ $$50=v_1\cdot t_1+v_2\cdot t_2\ $$ $$50=v_1\cdot t_1+\left(v_1+10\right)\cdot\left(t_1-\frac{1}{12}\right)\ $$ $$50=25+v_1\cdot t_1-\frac{v_1}{12}+10t_1-\frac{5}{6}$$ $$50=25+25-\frac{v_1}{12}+10t_1-\frac{5}{6}$$ $$\frac{5}{6}=-\frac{v_1}{12}+10t_1$$ $$10=-v_1+120t_1$$ $$\frac{10+v_1}{120}=t_1\ \ .$$ Now, replacing this in the equation d_1=v_1*t_1 we get $$d_1=v_1\cdot t_1\ \ \Rightarrow\ \ 25=v_1\left(\frac{10+v_1}{120}\right)\ \ \Rightarrow\ \ 3000=10v_1+v_1^2$$ $$v_1^2+10v_1-3000=0$$ $$(v_1-50)(v_1+60)=0$$ $$v_1=50\ \frac{km}{h}\ \ \ \ or\ \ \ \ v_1=-60\ \frac{km}{h}.$$
Thereore, the correct answer is the option C.

I hope it is clear enough.

Regards.

vaibhav101 wrote:A motorcyclist rode the first half of his way at a constant speed. Then he was delayed for 5 minutes and , therefore, to make for the lost time he increased his speed by 10 km/hr. Find the initial speed of the motorcyclist if the distance covered by him is equal to 50 km.

A 36 km/hr
B 48 km/hr
C 50 km/hr
D 62 km/hr
E 55 km/hr

We can PLUG IN THE ANSWERS, which represent the initial speed.
When the correct answer is plugged in, increasing the speed by 10 kph will enable the cyclist to travel 25 km in 5 fewer minutes.
Since the increase in the speed and the total distance are both multiples of 10, the correct answer is probably C, which also is a multiple of 10.

C: initial speed = 50 kph, higher speed = 50+10 = 60 kph
Time to travel 25 kilometers at a speed of 50 kph = d/r = 25/50 = 1/2 hour.
Time to travel 25 kilometers at a speed of 60 kph = d/r = 25/60 = 5/12 hour.
Time difference = 1/2 - 5/12 = 6/12 - 5/12 = 1/12 hour = 5 minutes.
Success!

The correct answer is C.