A car moves a distance of 200 m- It covers the first half of the distance speed 40 km-h and the second half of the distance speed v- If the average speed is 48 km-h- then the value of is-50 km-h56 km-h58 km-h60 km-h

Question

Toppr Admin · Accepted Answer

Given-xA0-Distance-200m-Speed-40km-h Average speed -48km-h48-20010040-100v-x21D2-140-1v-124-x21D2-v-1205-x2212-3-1202-x21D2-60km-h

A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of the distance at speed v. If the average speed is 48 km/h, then the value of v is-
50 km/h
56 km/h
58 km/h
60 km/h

Given, $D i s t a n c e = 200 m, S p e e d = 40 k m / h$ Average speed = $48 k m / h$

$48 = \frac{200}{\frac{100}{40} + \frac{100}{v}} \Rightarrow \frac{1}{40} + \frac{1}{v} = \frac{1}{24} \Rightarrow v = \frac{120}{5 - 3} = \frac{120}{2} \Rightarrow= 60 k m / h$

A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of the distance at speed v. If the average speed is 48 km/h, then the value of v is-50 km/h56 km/h58 km/h60 km/h

Given, Distance=200m,Speed=40km/h Average speed =48km/h48=20010040+100v⇒140+1v=124⇒v=1205−3=1202⇒=60km/h

A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of the distance at speed v. If the average speed is 48 km/h, then the value of v is-
50 km/h
56 km/h
58 km/h
60 km/h

Given, $D i s t a n c e = 200 m, S p e e d = 40 k m / h$ Average speed = $48 k m / h$

$48 = \frac{200}{\frac{100}{40} + \frac{100}{v}} \Rightarrow \frac{1}{40} + \frac{1}{v} = \frac{1}{24} \Rightarrow v = \frac{120}{5 - 3} = \frac{120}{2} \Rightarrow= 60 k m / h$