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?

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.

To find the speed at the first point, we can use the formula for constant acceleration:

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Given:
Distance (s) = 80 m
Time (t) = 7 s
Final velocity (v) = 15 m/s

We can calculate the acceleration first using the formula:
s = ut + 0.5at²

Given:
Distance (s) = 80 m
Time (t) = 7 s

Plug in the values into the formula and solve for acceleration (a):

80 = (u * 7) + 0.5 * a * (7^2)
80 = 7u + 24.5a

To solve this equation, we need another equation. We can use the equation v = u + at to calculate u (the initial velocity).

Given:
v = 15 m/s
t = 7s

15 = u + (7 * a)

Now we have a system of two equations with two variables:
Equation 1: 80 = 7u + 24.5a
Equation 2: 15 = u + (7 * a)

Solve this system of equations to find the values of u (initial velocity) and a (acceleration).

First, rearrange Equation 2 to find u in terms of a:

u = 15 - (7 * a)

Substitute this expression for u into Equation 1:

80 = 7(15 - 7a) + 24.5a
80 = 105 - 49a + 24.5a
80 = 105 - 24.5a
24.5a = 105 - 80
24.5a = 25
a = 25 / 24.5
a ≈ 1.02 m/s²

Substitute the value of a back into Equation 2 to find u:

15 = u + (7 * 1.02)
15 = u + 7.14
u ≈ 15 - 7.14
u ≈ 7.86 m/s

Hence, the speed at the first point is approximately 7.86 m/s, and the acceleration is approximately 1.02 m/s².