Question

During the first $18\min $ of a $60\min $ trip, a car Has an average speed of $11m{s^{ - 1}}$. What should be the average speed for the remaining 42min so that car has an average speed $21m{s^{ - 1}}$for the entire trip?
a. $25.3m{s^{ - 1}}$
b. $29.2m{s^{ - 1}}$
c. $31m{s^{ - 1}}$
d. $35.6m{s^{ - 1}}$

Answer 1

Hint: The entire trip can be divided into the part trip and their dimensions can be calculated separately. The average speed of the trip may be constant but the actual speed during different situations may be changing the simple speed distance relations will be used to find the average speed of the second part of the trip

Complete step by step answer:
Here in this question, it is given that the total time of the trip is $60\min $
 And the average speed during the entire trip is $21m{s^{ - 1}}$
So in other words the car is moving at $21m{s^{ - 1}}$for $60\min $
So total distance covered in this situation will be
$ d = s \times t$
$\Rightarrow d = 21 \times 60$
$ \Rightarrow d = 1260m$
Now the question says that during the first $18\min $of the trip car is moving at an average speed of $11m{s^{ - 1}}$
So the total distance covered in the case will be
$ d = s \times t$
$\Rightarrow d = 11 \times 18$
$ \Rightarrow d = 198m$
Now the total distance covered was $1260m$,
And the car is completing $198m$ in the first 18 minutes
So, the distance left to cover $1260 - 198 = 1062m$
And time remaining to cover this distance $42\min $
So, the average speed to cover this distance will be
$\because d = s \times t$
$\therefore s = \dfrac{d}{t}$
$ \Rightarrow s = \dfrac{{1062}}{{42}}$
$ \Rightarrow s = 25.285 \approx 25.3m{s^{ - 1}}$

Hence, the correct answer is option (A).

Note: The average speed during a trip is calculated as the total distance covered by the total time taken. The speed at some instances of the trip may be above or below depending upon the situation. If the car had a round trip and the velocity is asked then velocity should have been zero because the net displacement of the car will be zero.