Answer:
See solution below
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
