A racing car originally at rest, has a constant acceleration of 4m/s^2. When it reaches a max speed of 200km/h, it is subjected to a constant deceleration until in stops. If the distance travelled is 500m, find the deceleration and the total time elapsed. +
Need help with this question.
Originally posted by xunmeng:A racing car originally at rest, has a constant acceleration of 4m/s^2. When it reaches a max speed of 200km/h, it is subjected to a constant deceleration until in stops. If the distance travelled is 500m, find the deceleration and the total time elapsed. +
Need help with this question.
200km/h = 500/9 m/s
v^2 - u^2 = 2as
(500/9)^2 - 0 = 2(4)(s)
s = 385.8m
Amount of distance left before stopping
= 500m - 385.8m
= 114.2m
v^2 - u^2 = 2as
0 - (500/9)^2 = 2a(114.2)
a = -13.513ms^-2
s= ut + (1/2)at^2
114.2 = (500/9)t + 1/2(-13.513)t^2
-6.7566t^2 + (500/9)t - 114.2 = 0
t= [-b(+/-)sqrt(b^2-4ac)] / 2a, where a = -6.7566, b = (500/9), c = -114.2
t= 4.11s [ Note that the value of [sqrt(b^2-4ac) / 2a] is negligible in this case ]
Hi,
The key results for linear motion to be recalled:
1. v = u + at.
2. v^2 = u^2 + 2as.
3. s = ut + (1/2) at^2.
4. a > 0 (we have acceleration); a < 0 (we have deceleration); a = 0 (we have constant speed).
Thanks!
Cheers,
Wen Shih