Question

The motion of a body is given by the equation $\frac{dv}{dt} = 6.0 - 3v$, where $v$ is speed in m/s at time $t$ in sec. If body was at rest at $t=0$

• A. The terminal speed is 2.0 m/s
• B. The speed varies with the time as $v(t) = 2(1-e^{-3t})$ m/s
• C. The speed is 0.1 m/s when the acceleration is half the initial value
• D. The magnitude of the initial acceleration is 6.0 m/s²

Answer 1

Answer: (A), (B), (D)

A, B, D

Answer 2

The motion of the body is described by the differential equation: $\frac{dv}{dt} = 6.0 - 3v$

Given initial condition: at $t=0$, $v=0$.

Let's analyze each option:

A) The terminal speed is 2.0 m/s

Terminal speed ($v_t$) is reached when the acceleration becomes zero, i.e., $\frac{dv}{dt} = 0$.

Setting the given equation to zero:

$6.0 - 3v_t = 0$

$3v_t = 6.0$

$v_t = \frac{6.0}{3} = 2.0$ m/s

Thus, option (A) is correct.

B) The speed varies with the time as $v(t) = 2(1-e^{-3t})$ m/s

To find the speed as a function of time, we need to solve the differential equation:

$\frac{dv}{dt} = 6 - 3v$

Separate the variables:

$\frac{dv}{6 - 3v} = dt$

Integrate both sides. To make integration easier, factor out -3 from the denominator:

$\frac{dv}{-3(v - 2)} = dt$

$\int \frac{dv}{v - 2} = \int -3 dt$

$\ln|v - 2| = -3t + C$

$v - 2 = A e^{-3t}$ (where $A = \pm e^C$)

Now, apply the initial condition: at $t=0$, $v=0$.

$0 - 2 = A e^{-3(0)}$

$-2 = A \cdot 1$

$A = -2$

Substitute $A = -2$ back into the equation:

$v - 2 = -2 e^{-3t}$

$v(t) = 2 - 2 e^{-3t}$

$v(t) = 2(1 - e^{-3t})$ m/s

Thus, option (B) is correct.

C) The speed is 0.1 m/s when the acceleration is half the initial value

First, calculate the initial acceleration ($a_{initial}$). Initial acceleration occurs at $t=0$, when $v=0$.

$a_{initial} = \left(\frac{dv}{dt}\right)_{t=0} = 6.0 - 3(0) = 6.0$ m/s²

Half the initial acceleration is $\frac{6.0}{2} = 3.0$ m/s².

Now, find the speed ($v$) when the acceleration is $3.0$ m/s²:

$3. 0 = 6.0 - 3v$

$3v = 6.0 - 3.0$

$3v = 3.0$

$v = 1.0$ m/s

The option states the speed is 0.1 m/s, which is incorrect.

Thus, option (C) is incorrect.

D) The magnitude of the initial acceleration is 6.0 m/s²

As calculated in option (C), the initial acceleration is when $t=0$ (and $v=0$):

$a_{initial} = 6.0 - 3(0) = 6.0$ m/s²

Thus, option (D) is correct.

Final Answer: The correct options are (A), (B), and (D).