Question

A physics instructor drives down to St. Louis to see his mom. On the way there, he travels the first half of the time at 65 km/hr and the second half the time at 95 km/hr. Coming back, he drives the first half the distance at 65 km/hr and the second half of the distance at 95 km/hr.
What is his average speed going down to St. Louis?
What is his average speed coming back from St. Louis?
What is his average speed during the entire trip?
What is his total displacement during the entire trip?
Which of these above graphs best represents his velocity during the trip to St. Louis?
Put in the number below the correct graph Which of these above graphs best represents his displacement during

Answer 1

Answer:

$v_{avg}=77.1874\ km.hr^{-1}$

$v'_{avg}=77.1874\ km.hr^{-1}$

$v_a=77.1874\ km.hr^{-1}$

$s=0\ km$ because the person finally return to the initial position.

Explanation:

Given:

velocity during the first half, $u_1=65\ km.hr^{-1}$

velocity during the second half, $u_2=95\ km.hr^{-1}$

velocity during the first half of onward journey, $v_1=65\ km.hr^{-1}$

velocity during the second half of onward journey, $v_2=95\ km.hr^{-1}$

Let the total displacement and distance between the two points be 2x meters.

So, the time taken for the first half of the journey:

$t_1=\frac{x}{65} \ hr$

The time taken for the second half of the journey:

$t_2=\frac{x}{95} \ hr$

Time taken for the first half of onward journey:

$t'_1=\frac{x}{65} \ hr$

Time taken for the first half of onward journey:

$t'_2=\frac{x}{95}\ hr$

average speed going down to St. Louis:

$v_{avg}=\frac{Total\ distance}{total\ time}$

$v_{avg}=\frac{2x}{t_1+t_2}$

$v_{avg}=\frac{2x}{(\frac{x}{65}+\frac{x}{95}) }$

$v_{avg}=77.1874\ km.hr^{-1}$

Similarly, average speed coming back from St. Louis:

$v'_{avg}=\frac{2x}{t'_1+t'_2}$

$v'_{avg}=\frac{2x}{(\frac{x}{65}+\frac{x}{95}) }$

$v'_{avg}=77.1874\ km.hr^{-1}$

Now the total average speed of the trip:

$v_a=\frac{2x+2x}{(t_1+t_2+t'_1+t'_2)}$

$v_a=\frac{4x}{(\frac{x}{65} +\frac{x}{95} +\frac{x}{65} +\frac{x}{95} )}$

$v_a=77.1874\ km.hr^{-1}$

Total displacement during the entire trip:

$s=0\ km$ because the person finally return to the initial position.