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Question: A car drives north for \(35\) minutes at \(85\,kmh{r^{ - 1}}\) and then stops for \(15\min .\) The c...

A car drives north for 3535 minutes at 85kmhr185\,kmh{r^{ - 1}} and then stops for 15min.15\min . The car continues north travelling 130km130\,km in 22 hours. What is the total displacement and average velocity?

Explanation

Solution

In order to solve this question, we should know that displacement covered by the body is distance between final and initial point of path followed by the body and average velocity is the ratio of net displacement and total time taken by the body to cover that displacement, here we will use given information to find displacement and average velocity of the car.

Complete step by step answer:
According to the question, we have given that car travels in one direction north and let d1{d_1} be the distance covered by car in t=35min=712hrt = 35\min = \dfrac{7}{{12}}hr with a velocity of v=85kmhr1v = 85\,kmh{r^{ - 1}} so, using distance formula we get,
d1=v×t{d_1} = v \times t
On putting value of parameters we get,
d1=85×712{d_1} = 85 \times \dfrac{7}{{12}}
d1=49.6km\Rightarrow {d_1} = 49.6km

Now, a car stops for t=15mint = 15\min in which it covers zero distance and then, car covers a distance of d2=130km{d_2} = 130km for time period of t=2hrt = 2hr so, Net displacement covered by the car D is,
D=d1+d2D = {d_1} + {d_2}
On putting the value of parameters we get,
D=130+49.6D = 130 + 49.6
D=179.6km\Rightarrow D = 179.6\,km

For average velocity we have, net displacement by car is D=179.6kmD = 179.6\,km and total time taken by the car T is
T=712hr+2hr+15minT = \dfrac{7}{{12}}hr + 2hr + 15\min
Or we can write it as,
T=0.58+2+0.25=2.83hrT = 0.58 + 2 + 0.25 = 2.83hr
And average velocity VV is,
V=DT\Rightarrow V = \dfrac{D}{T}
On putting the value of parameters we get,
V=179.62.83 V=63.46kmhr1V = \dfrac{{179.6}}{{2.83}} \\\ \therefore V= 63.46\,kmh{r^{ - 1}}

Hence, the displacement covered by the car is 179.6km179.6\,km and the average velocity of the car is 63.46kmhr163.46\,kmh{r^{ - 1}}.

Note: It should be remembered that, basic conversion of time from minutes into hour is used here as 1min=160hr1\min = \dfrac{1}{{60}}hr and car covers the distance in one direction only so here, displacement and distance have the same magnitude but displacement is a vector quantity and here the direction of displacement is in North.