Problem:
The radius vector of a point depends on time t, as r = ct { (bt²)/2} where c and b are constant vectors. Find the modulus of velocity and acceleration at time t.

Solution:
1. Modulus of Velocity:
The velocity of a particle is given by the derivative of the position vector with respect to time. To find the modulus of velocity, we need to find the magnitude of the velocity vector.

Given: r = ct { (bt²)/2}

Differentiating r with respect to time t, we get:

v = dr/dt = c + (bt) * b

The magnitude of the velocity vector v is given by:

|v| = sqrt((c + (bt) * b) * (c + (bt) * b))

Simplifying the expression, we get:

|v| = sqrt(c * c + 2 * c * (bt) * b + (bt * b) * (bt * b))

|v| = sqrt(c * c + 2 * c * (bt) * b + (bt)² * (b * b))

|v| = sqrt(c * c + 2 * c * (bt) * b + (bt)² * |b|²)

2. Modulus of Acceleration:
The acceleration of a particle is given by the derivative of the velocity vector with respect to time. To find the modulus of acceleration, we need to find the magnitude of the acceleration vector.

Differentiating v with respect to time t, we get:

a = dv/dt = b * b

The magnitude of the acceleration vector a is given by:

|a| = sqrt((b * b) * (b * b))

|a| = sqrt(|b|² * |b|²)

|a| = sqrt(|b|⁴)

Summary:
To find the modulus of velocity, differentiate the position vector with respect to time, find the magnitude of the resulting velocity vector, and simplify the expression. To find the modulus of acceleration, differentiate the velocity vector with respect to time, find the magnitude of the resulting acceleration vector, and simplify the expression.

The modulus of velocity is given by |v| = sqrt(c * c + 2 * c * (bt) * b + (bt)² * |b|²).

The modulus of acceleration is given by |a| = sqrt(|b|⁴).