Question

A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and second half of the distance at speed v. The average speed is 48 km/h. The value of v is
(1) 56 km/h
(2) 60 km/h
(3) 50 km/h
(4) 58 km/h​

Answer 1

Given :-

◉ A car moves a distance of 200 m.

◉ The car covers the first half of the distance at speed 40 km/h and second half of the distance at speed v.

◉ Average speed is 48km/h

To Find :-

◉ Value of v

Solution :-

Since we are given the total distance travelled hy the car i.e., 200 m. So, we just have to find the total time taken.

Case 1

It is given that the car covered half of the distance with speed 40 km/h

We have,

• Speed = 40 km/h
• Distance = 200/2 ⇒ 100 m or 0.1 Km

We know,

⇒ Distance = Speed × Time

⇒ 0.1 = 40 × t₁

⇒ t₁ = 1/400 ...(1)

Case 2

Also, The car covered next half of the distance with speed v, we have

• Distance travelled = 100 m or 0.1 km
• Speed = v km/h

We know,

⇒ Distance = Speed × Time

⇒ 0.1 = v × t₂

⇒ t₂ = 1/10v ...(2)

Now, It is given that the average speed of the car is 48 km /hr , we have

• Total distance travelled = 200 m or 0.2 km
• Total time taken = t₁ + t₂

⇒ Average speed = Total distance / Total time

⇒ 48 = 0.2 / (t₁ + t₂)

⇒ 48 / 0.2 = 1 / (1/400 + 1/10v)

⇒ 240 = 1 / { (v + 40)/400v }

⇒ 240 = 400v / (v + 40)

⇒ 240v + 9600 = 400v

⇒ 160v = 9600

⇒ v = 60 km/h

Hence, The value of v is 60 km/h

∴ Option (2) is correct.

Answer 2

$\star\:\:\:\bf\large\underline\blue{Given:—}$

Distance (d) = 200 m

Speed = 40 km/h

Average speed (s) = 48 km/h

$\star\:\:\:\bf\large\underline\blue{To\:find:—}$

Value of ' v '

$\star\:\:\:\bf\large\underline\blue{Solution:—}$

We know that,

$\boxed{\bf{\red{s=\frac{d}{t}}}}$

$➞48 = \frac{200}{ \frac{100}{40} + \frac{100}{v} } \\ \\➞ \frac{1}{40} + \frac{1}{v } = \frac{1}{24} \\ \\ ➞ v = \frac{120}{5 - 3} \\ \\ ➞v = \frac{120}{2} \\ \\ ➞ v = 60$

Hence, the value of v is 60 km/h