A particle moves along a straight line It starts with an ini

A particle moves along a straight line. It starts with an initial velocity of 1 m/s and continue with acceleration a=12t for 0.2 seconds. Then it decelerates at a=-3v^-3(m/s²) until it stops. Determine how far the particle has traveled during the entire process.

Solution

Initial velocity v₀ = 1 m/s

Let velocity at t = 0.2 s be v_0.2

a = dv/dt = 12t

dv = 12t dt

Integrating, v = 6t² + C

At t = 0, we have v = 1.

Thus, 1 = 0 + C

C = 1

v = 6t² + 1

At t = 0.2 we get,

v_0.2 = 1.24 m/s

v = ds/dt

ds = (6t² + 1) dt

Integrating s = 2t³ + t + C

At t = 0, we have s = 0

Thus, 0 = 0 + 0 + C

C = 0

Therefore, s = 2t³ + t

At t= 0.2 s, we get

s_0.2 = 2*0.2³ + 0.2

s_0.2 = 0.216 m

During the deceleration portion, a = -3v^-3

a = dv/dt

dv/dt = (-3v^-3)

dv / (-3v^-3) = dt

dv / (v^-3) = -3 dt

Integrating both the sides,

v⁴ /4 = -3t + C

At t = 0.2 s, we have v = 1.24 m/s

1.24⁴ /4 = -3*0.2 + C

C = 1.191

Thus, v⁴ /4 = -3t + 1.191

v⁴ = -12t + 4.7642

For v = 0, we\'ll have 0 = -12t + 4.7642

or t = 0.397

v = ds/dt

ds = [-12t + 4.7642]^0.25 dt

Integrating both the sides,

s = -[-12t + 4.7642]^1.25 / (1.25*12) + C

At t = 0.2 s, we have s = 0.216 m

0.216 = -[-12*0.2 + 4.7642]^1.25 / (1.25*12) + C

C = 0.4114

Thus, s = -[-12t + 4.7642]^1.25 / (1.25*12) + 0.4114

At t = 0.397 s we get

s = -[-12*0.397 + 4.7642]^1.25 / (1.25*12) +  0.4114

s = 0.41139 m