Question

Two cars are initially moving with speeds vA and vB. The cars are decelerated at the same rate until they come to a stop. If it takes car A four times as far to stop as car B, then how do their initial speeds compare?

Answer 1

`v_a = 2 v_b`

Step-by-step explanation:

As we know that the speed of car A and car B is given by

`v_a` &
`v_b`

now we know that both cars are decelerated by same deceleration and stopped finally

so the distance moved by the car is given by the equations

`v_f^2 - v_i^2 = 2a d`

`0 - v_a^2 = 2(-a) d_a`

`d_a = (v_a^2)/(2a)`

similarly we have

`d_b = (v_b^2)/(2a)`

now we know that

`d_a = 4 d_b`

`(v_a^2)/(2a) = 4 (v_b^2)/(2a)`

`v_a = 2 v_b`