Question

The velocity of a particle is described by an equation v=ct +dt-2 where c=0.1m/s and d=0.02m/s Calculate

Answer 1

The equation v=ct+dt-2 can be used to calculate the velocity of a particle given specific values for c and d.

Step-by-step explanation:

The equation given is v = ct + dt - 2, where c = 0.1 m/s and d = 0.02 m/s. To calculate the velocity of the particle, substitute the values of c and d into the equation and simplify:

v = (0.1t) + (0.02t) - 2

For example, if t = 2 seconds:

v = (0.1 * 2) + (0.02 * 2) - 2

v = 0.2 + 0.04 - 2

v = -1.76 m/s

Answer 2

See solution below

Step-by-step explanation:

Complete question

The velocity of a particle is described by an equation v= ct + dt²

where c = 0.1 m/s and d = 0.02 m/s. Calculate:

(i) the change in velocity during the time interval t = 3 seconds and

t = 6 seconds.

(ii) the average acceleration in that same time interval

Substitute the value of c and d into the equation to have

v= ct + dt²

v = 0.1t+0.02t²

The velocity at the = 3secs will be expressed as:

V(3) = 0.1(3)+0.03(3)²

v(3) = 0.3+0.27

V(3) = 0.57m/s

When t =6secs

V(6) = 0.1(6)+0.03(6)²

v(6) = 0.6+1.08

V(6) = 1.68m/s

Change in velocity = v(6)-v(3)

Change in velocity = 1.68-0.57

Change in velocity= 1.11m/s

B) Acceleration is the rate of change in velocity

A(t) = dv/dt

A(t) = c+2dt

A(t) = 0.1+2(0.02)t

A(t) = 0.1+0.04t

If t = 3

A(3) = 0.1+0.04(3)

A(3)= 0.1+0.12

A(3) = 0.22m/s²

A(6) = 0.1+0.04(6)

A(6)= 0.1+0.24

A(6) = 0.34m/s²

Average acceleration = 0.34-0.22

Average acceleration = 0.12m/s²