Question

Dave wishes to get produce from the store, which is a distance D- 2.2 km east of his house. Dave drives his bike to the store at a constant speed of V,-6.5 m/s, and drives back to his house at the slower speed of v2 = 2.1 m/s. House Otheexpertta.con Part (a) How long (in seconds) does it take for Dave to reach the store? Numerie A numeric value is expected and not an expression. t= Part (b) In seconds, how long does the whole trip take? Numeric : A numeric value is expected and not an expression. Part (c) How long does the trip take in minutes? Numeric : A numeric value is expected and not an expression. minutes Part (d) How much distance, in kilometers, did Dave travel during the whole trip? Numerie : A numeric value is expected and not an expression. Part (e) What is the magnitude of Dave's displacement, in km, for the entire trip?

Answer 1

Answer:

a) $t_1=338.4615\ s$

b) $t=1386.0805\ s$

c) $t=23.1013\ min$

d) $d=4400\ m$

e) Since Dave starts from house and finally returns to the house so displacement is zero.

Explanation:

Given:

• distance between the house and the store, $s=2.2\ km=2200\ m$

• speed of driving from house to store, $v_1=-6.5\ m.s^{-1}$

• speed of driving back from store to house, $v_2=2.1\ m.s^{-1}$

Since the store is located towards east from his house and the velocity in this direction is taken negative and contrary to this the velocity in the west direction is taken positive.

a)

time taken in reaching the store:

$t_1=\frac{s}{v_1}$

$t_1=\frac{-2200}{-6.5}$

$t_1=338.4615\ s$

b)

Now the time taken in returning form the store:

$t_2=\frac{s}{v_2}$

$t_2=\frac{2200}{2.1}$

$t_2=1047.6190\ s$

therefore total time taken by the trip:

$t=t_1+t_2$

$t=338.4615+1047.6190$

$t=1386.0805\ s$

c)

the time taken by the trip in minutes:

$t=\frac{1386.0805}{60}$

$t=23.1013\ min$

d)

distance travelled in the whole trip:

$d=2\times s$

$d=2\times 2200$

$d=4400\ m$

e)

Since Dave starts from house and finally returns to the house so displacement is zero.