Question

A car’s brakes decelerate it at a rate of -2.20 m/s2. If the car is originally travelling at 17 m/s and comes to a stop, then how far, in meters, will the car travel during that time?

Answer 1

A :-) for this question , we should apply
2as = v^2 - u ^2
Given - u = 17 m/s
v = 0 m/s
a = -2.20 m/s^2
Solution -
2as = v^2 - u^2
2 x -2.20 x s = 0^2 - 17^2
s = 0^2 - 17^2 by 2 x -2.20
s = -( 17 x 17 ) by 2 x -2.20
( Cut minus ( - ) and ( - ) )
s = 17 x 17 by 2 x 2.20
s = 289 by 4.4
s = 65.6

.:. The car travelled 65.6 meters.

Answer 2

x = 65.68 [m]

Step-by-step explanation:

To solve this problem we must use the following equation of kinematics.

`v_(f)^(2)=v_(o)^(2)-2*a*x`

where:

Vf = final velocity = 0 (comes to stop)

Vo = initial velocity = 17 [m/s]

a = desaceleration = -2.2 [m/s²]

x = displacement [m]

Now replacing, we have:

`(0)^(2)=(17)^(2)-2*(2.2)*x\\4.4*x = 289\\x = 65.68 [m]`