Two cars which are 40 miles apart start moving simultaneously in the same direction with constant speeds. if the speed of the car which is behind is 56 mph and the speed of the other car is 48 mph, how many hours will it take for the car which is behind to catch up the car ahead of it?

5mph

Step-by-step explanation:

Given that two cars are 40 miles apart at the beginning.

They started at the same time with speeds 48 mph and 56 mph respectively

Since both cars are travelling in the same direction,

Relative velocity = Speed of the behind car-Speed of the first car

=56-48 =8mph

To catch the I car, the second car has to travel a distance of 40 miles extra in the same time as the I car

i.e. speed to be made up by the second car = 40 miles

Relative velocity per hour              = 8mph

Hence time taken to catch up =

x = 25/2304

step-by-step explanation:

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12