Question

The motion of a body is given by the equation a=4-2v, where v is speed in m/s, a is acceleration, then the speed of body as a function of time is​

Answer 1

The motion of a body is given by the equation, a = 4 - 2v

we know, acceleration is the rate of change of velocity with respect to time.

so, a = dv/dt = 4 - 2v

⇒∫dv/(4 - 2v) = ∫dt

⇒(1/2) ∫dv/(2 - v) = ∫dt

⇒-(1/2)ln(2 - v) = t + c , where c is constant.

⇒ln(2 - v) = -2(t + c)

⇒2 - v = $e^{-2(t + c)}$

⇒2 - $e^{-2(t + c)}$ = v

⇒v = 2 - $e^{-2t}e^{-2c}$

we know, e^2c is also a constant let e^(-2c) = k

so, v = $2-ke^{2t}$

hence, speed of body as a function of time is given as $2-ke^{2t}$

Answer 2

Answer:

Explanation:

Given that, dvdt=4−2v

(i) At t = 0, v = 0, so initial acceleration is 4−0=4m/s2

(ii) dv=(4−2v)dt

or,dv4−2v=dt

or, dv=∫v0dv4−2v=∫t0dt

loge(4−2v)−2:∣v0=(t)t0=t

loge{4−2v}−loge(4)=−2t

loge{4−2v4}=−2t

⇒4−2v4=e−2t⇒44−2v4=e−2t

⇒1−2v4=e−2t

⇒2v4=1−e−2t

or v=2(1−e−2t)