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

b would be the answer