Question

A car moving at a constant acceleration covers a distance within two point that are 80m apart in 7s, when it's speed at the 2nd point is 15m/s . what it is speed at the first point? Also what is the acceleration?

Answer 1

let the acceleration be a m/s^2
let the velocity be v m/s
let the distance be s m

v = at + c
when t = 7, v = 15
15 = 7a + c -----> c = 15-7a **
when t = 0, ....
v = c

s = (1/2)at^2 + t(15-7a) + k

when t = 0, s = 0 + 0 + k = k
when t = 7, s = (49/2)a + 105 - 48a + k

(49/2)a + 105 - 48a + k - k = 80
49a/2 - 48a = -25
(-47/2)a = -25
a = 50/47
back in **
c = 15-7a
= 15-7(50/47) = 355/47

at the first point: t = 0
v = at + c
= c = 355/47 m/s or appr 7.55 m/s

a = 50/47 m/s^2 or appr 1.064 m/s^2


checking:

a = 1.064
v = 1.064t + 7.55
when t = 0, v = 7.55
when t = 7, v = 1.064(7) + 7.55 = 14.998 , not bad

when t = 0 , s = k
when t = 7,
s = (1/2)(1.064)(49) + 7(15-1.064(7)) + k
= 78.932 + k

difference = 78.932+k - k = 78.932 or appr 79 m
off by 1 m, due to rounding of decimals.