The acceleration of a particle is given as a = kt2. Where k is a constant coefficient and t is time in seconds. At time t = 0, x = 24 m; At t = 6 seconds, it is x = 96 m and V = 18 m / s. Obtain the path (x) and velocity (V) equations depending on time.

s = t⁴/108+6t+24

v = t³/27 + 6

Explanation:

Given the equation of acceleration expressed as a = kt²

t is in seconds

To get the velocity, we will integrate the acceleration function

a = kt²

Integrate both sides

a = kt²

v = ∫kt² dt

v = k ∫t² dt

v = kt³/3 + C1 *

s = ∫kt³/3 dt+ ∫C1 dt

s = kt⁴/12+C1t + C2

At t = 0, S= 24

24 = k(0)⁴/12 + C1(0)+ C2

24 = C2

s = kt⁴/12+C1t + 24

If at t = 6 seconds, s = 96m

Substitute;

96 = k6⁴/12+6C1+ 24 **

Also at t = 6, v = 18m/s

Substitute into *

v = kt³/3 + C1..

18 = k6³/3 + C1 ***

Solve ** and *** simultaneously and get k and C1

96 = k6⁴/12+6C1+ 24

k6⁴/12+6C1+ = 72 × 1

k6³/3 + C1 = 18 × 6

k6⁴/12+6C1+ = 72

k6⁴/3 + 6C1 = 108

Subtract

[(6⁴k)-4(6⁴k)]/12 = 72-108

-3(6⁴k)/12 = -36

6⁴k/12 = 12

6⁴k = 144

k = 144/1296

k = 1/9

Substitute k = 1/6 into k6³/3 + C1 = 18 and get C1

1/6(216/3) + C1 = 18

12+ C1= 18

C1 = 18-12

C1 = 6

Write s in terms of t;

s = kt⁴/12+C1t + 24

s = 1/9t⁴/12+6t+24

s = t⁴/108 + 6t + 24

Write v in terms of t;

v = kt³/3 + C1

v = 1/9(t³/3)+6

v = t³/27 + 6

Step-by-step explanation:

The given equation is

In this equation, the property applied was the identity property for addition, which states that zero added to any number is the number itself, that is

This identity property is using the neutral element for addition which is the zero, because it doesn't increase or decrease the number that's being operated with.

Therefore, the right answer is C) identity property.