Question

particle’s velocity is described by the function vx=kt2
m/s, where k is a constant and t is in s. The particle’s position
at t0=0 s is x0= –9.0 m. At t1=3.0 s, the particle is at x1=9.0 m. Determine the value of the constant k.

Answer 1

Explanation:

The value of the constant k is 2

Explanation:

We have the equation of the velocity v = kt², where k

is constant and t is the time in second

The particle's position at to = 0 is xo = -9 m

The particle's position at t₁ = 3 s is x₁ = 9 m

We need to find the value of the constant k

The relation between the velocity and the

displacement in a particular

time is x = f va dt

Remember in integration we add power by 1 and divide the expression

by the new power

⇒ x = S kt² dt = }, kt³ + c

c is the constant of integration to find it substitute the initial value of x

and t in the equation of x

to = 0, xo = -9 m

-9 = ¹ k (0)³ + c

→ 9 = ¹ (27) k - 9

3

-9=9k-9

Add 9 to both sides

-18-9 k

Divide both sides by 9

→ k = 2

The value of the constant k is 2