Solve the problem.
-Suppose that the velocity of a falling body is $\mathrm { v } = \mathrm { ks } ^ { 2 }$ ( $\mathrm { k }$ a constant) at the instant the body has fallen s meters from its starting point. Find the body's acceleration as a function of s.
A) $a = 2 k s ^ { 2 }$
B) $a = 2 \mathrm { ks }$
C) $a = 2 k ^ { 2 } s ^ { 3 }$
D) $a = 2 k s ^ { 3 }$

Question

Quizplus · Accepted Answer

Acceleration is the derivative of velocity with respect to time. Given $v = ks^2$, we use the chain rule to find acceleration $a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt}$. Since $v = ks^2$, $\frac{dv}{ds} = 2ks$, and $\frac{ds}{dt} = v = ks^2$, thus $a = 2ks \cdot ks^2 = 2ks$.

Solve the Problem v=ks2\mathrm { v } = \mathrm { ks } ^ { 2 }v=ks2

Solve the Problem $\mathrm { v } = \mathrm { ks } ^ { 2 }$