Question
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 35 minutes at 85kmhr−1 and then stops for 15min. The car continues north travelling 130km in 2 hours. What is the total displacement and average velocity?
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 be the distance covered by car in t=35min=127hr with a velocity of v=85kmhr−1 so, using distance formula we get,
d1=v×t
On putting value of parameters we get,
d1=85×127
⇒d1=49.6km
Now, a car stops for t=15min in which it covers zero distance and then, car covers a distance of d2=130km for time period of t=2hr so, Net displacement covered by the car D is,
D=d1+d2
On putting the value of parameters we get,
D=130+49.6
⇒D=179.6km
For average velocity we have, net displacement by car is D=179.6km and total time taken by the car T is
T=127hr+2hr+15min
Or we can write it as,
T=0.58+2+0.25=2.83hr
And average velocity V is,
⇒V=TD
On putting the value of parameters we get,
V=2.83179.6 ∴V=63.46kmhr−1
Hence, the displacement covered by the car is 179.6km and the average velocity of the car is 63.46kmhr−1.
Note: It should be remembered that, basic conversion of time from minutes into hour is used here as 1min=601hr 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.