- #1
hellothere.
- 2
- 0
Sorry about this, must seem a trivial question, but anyways here goes.
A car is traveling at 13ms^1. It is allowed to slow down natrually without appyling the brakes. The distance traveled in coming to rest is 640m.
Show that the average deceleration of the car is about 0.1ms^2.
-------------------
Ok so first of since time aint metioned i used.
v^2 = u^2 + 2as
giving me
v = 0
u =13
a = ?
s = 640
----------------------------------
v^2 = 13^2 + 2(a)640.
v^2= 13^2 +2(640)
___________
a
a * v^2 = 13^2 +2(640)
a = 13^2 +2(640)
__________
v^2
Giving me a very strange answer, no were remotely close to 0.1, I've obviously screwed up the re-arranging or used the wrong formula.
Would like a nudge in the right direction please.
A car is traveling at 13ms^1. It is allowed to slow down natrually without appyling the brakes. The distance traveled in coming to rest is 640m.
Show that the average deceleration of the car is about 0.1ms^2.
-------------------
Ok so first of since time aint metioned i used.
v^2 = u^2 + 2as
giving me
v = 0
u =13
a = ?
s = 640
----------------------------------
v^2 = 13^2 + 2(a)640.
v^2= 13^2 +2(640)
___________
a
a * v^2 = 13^2 +2(640)
a = 13^2 +2(640)
__________
v^2
Giving me a very strange answer, no were remotely close to 0.1, I've obviously screwed up the re-arranging or used the wrong formula.
Would like a nudge in the right direction please.