Question

Calculus method of solving the kinematics equation

Answer 1

Answer:

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Step-by-step explanation:

Calculus is an advanced math topic, but it makes deriving two of the three equations of motion much simpler. By definition, acceleration is the first derivative of velocity with respect to time. Take the operation in that definition and reverse it. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. This gives us the velocity-time equation. If we assume acceleration is constant, we get the so-called first equation of motion [1].

a =

dv

dt

dv = a dt

v

⌠

⌡ dv

v0 =

t

⌠

⌡ a dt

0

v − v0 = at

v = v0 + at [1]

Answer 2

Answer:

Derivation of all the three kinematic equations by calculus method -

1. We know, Acceleration = Δv/Δt = dv/dt

$\longrightarrow \tt{dv = a\,dt}$

$\longrightarrow \tt{\int\limits^v_u dv= \int\limits^t_0 a \, dt}$

$\longrightarrow \tt{[v]^v_u= a\int\limits^t_0 dt}$

$\longrightarrow \tt{[v]^v_u= a[t]^t_0}$

$\longrightarrow \tt{v - u = a(t - 0)$

$\longrightarrow \tt{v - u = at}$

Hence, v = u + at.

2. We know that, Velocity = Δs/Δt = ds/dt

$\longrightarrow \tt{ds = v\,dt}$

$\longrightarrow \tt{\int\limits^s_0 ds= \int\limits^t_0 v \, dt}$

$\longrightarrow \tt{[s]^s_0= \int\limits^t_0(u + at)\,dt}$

$\longrightarrow \tt{[s]^s_0= \int\limits^t_0 udt + atdt$

$\longrightarrow \tt{[s]^s_0= u\int\limits^t_0 dt + a\int\limits^t_0 tdt}$

$\longrightarrow \tt{[s]^s_0= u[t]^t_0 + a\bigg[\dfrac{{t}^2}{2}\bigg]^t_0}$

$\sf\bigg[By\,using : \int\limits {x}^n \, dx = \dfrac{{x}^{n+1}}{n+1}\bigg]$

$\longrightarrow \tt{s - 0 = u(t - 0) + a\bigg(\dfrac{t^{2}}{2}-\dfrac{0^{2}}{2}\bigg)$

$\longrightarrow \tt{s - 0 = u(t - 0) + at^2}$

Hence, s = ut + 1/2 at².

3. Acceleration = dv/dt

Multiplying ds in numerator and denominator,

$\longrightarrow \tt{a=\dfrac{dv}{dt}\times\dfrac{ds}{ds}}$

$\longrightarrow \tt{a=\dfrac{dv}{ds}\times\dfrac{ds}{dt}}$

$\longrightarrow \tt{a=v\dfrac{dv}{ds}}}$

$\longrightarrow \tt{a\,ds = v\,dv}}$

$\longrightarrow \tt{\int\limits^s_0 {a} \, ds = \int\limits^v_u {v} \, dv}$

$\longrightarrow \tt{\int\limits^s_0 {a} \, ds = \bigg[\dfrac{v^2}{2}\bigg]^v_u}$

$\sf\bigg[By\,using : \int\limits {x}^n \, dx = \dfrac{{x}^{n+1}}{n+1}\bigg]$

$\longrightarrow\tt{a[s]^s_0=\dfrac{v^2}{2}-\dfrac{u^2}{2}}$

$\longrightarrow\tt{a(s-0)=\dfrac{v^2-u^2}{2}}$

$\longrightarrow\tt{2a(s-0)=v^2-u^2}$

Hence, 2as = v² - u².