Answer:
1001 m/s
300 [tex]m/s^2[/tex]
0.299.
Step-by-step explanation:
Given that:
[tex]D=f(t)=t^3+1[/tex]
[tex]f(10)[/tex] means putting the value of [tex]t=10[/tex] in [tex]f(t)[/tex].
[tex]f(10) = 10^3+1=1001[/tex] [tex]m/s[/tex]
To find [tex]f'(10)[/tex], we need to find [tex]f'(t)[/tex] first.
Differentiating the equation of Distance [tex]D[/tex] w.r.to [tex]t[/tex]:
[tex]f'(t) = \dfrac{d}{dt}(t^3+1)= 3t^{3-1}+1 = 3t^2[/tex]
[tex]f'(10) = 3(10)^2=300[/tex] [tex]m/s^2[/tex]
Relative rate of change:
[tex]\dfrac{f'}{f}\ at\ t = 10 \Rightarrow \dfrac{300}{1001} \approx 0.299[/tex]
