Question

A body of mass 2 kg is moving in straight line for which velocity-position (v-x) graph is shown in the figure. At t = 0, body is at origin

• A. Force acting on the body at x = 10 m is 20 N
• B. x = 20(1-e⁻ᵗ) m
• C. Acceleration of body is uniform
• D. Acceleration-position graph is parabola

Answer 1

Answer: (A), (B)

A, B

Answer 2

The velocity-position (v-x) graph is a straight line passing through (0, 20) and (20, 0). The equation of this line is $v(x) = 20 - x$.

The acceleration of the body is given by $a = v \frac{dv}{dx}$. From $v(x) = 20 - x$, we have $\frac{dv}{dx} = -1$. Therefore, $a = (20 - x)(-1) = x - 20$. The mass of the body is $m = 2$ kg. The force acting on the body is $F = ma = 2(x - 20)$.

Analysis of options:

A. Force acting on the body at x = 10 m is 20 N At $x = 10$ m, the acceleration is $a = 10 - 20 = -10$ m/s². The force is $F = ma = (2 \text{ kg})(-10 \text{ m/s}^2) = -20$ N. The magnitude of the force is $|-20|$ N = 20 N. Assuming the question refers to the magnitude of the force, this statement is correct.

B. x = 20(1 - e⁻ᵗ) m We have $a = \frac{dv}{dt}$. So, $\frac{dv}{dt} = x - 20$. Also, $v = \frac{dx}{dt}$. From $v = 20 - x$, we get $\frac{dx}{dt} = 20 - x$. This is a first-order linear differential equation: $\frac{dx}{dt} = 20 - x$. Separating variables: $\frac{dx}{20 - x} = dt$. Integrating both sides: $\int \frac{dx}{20 - x} = \int dt$. This gives $-\ln|20 - x| = t + C$. Using the initial condition that at $t = 0$, $x = 0$, we get $-\ln|20 - 0| = 0 + C$, so $C = -\ln(20)$. Substituting C back: $-\ln|20 - x| = t - \ln(20)$. Rearranging: $\ln|20 - x| = \ln(20) - t$. Exponentiating both sides: $|20 - x| = e^{\ln(20) - t} = e^{\ln(20)} e^{-t} = 20 e^{-t}$. Since $v = 20 - x$ and $v(0) = 20$ m/s, and $v$ decreases as $x$ increases, $v$ is positive for $x < 20$. Thus, $20 - x > 0$. So, $20 - x = 20 e^{-t}$. This yields $x = 20 - 20 e^{-t} = 20(1 - e^{-t})$. This statement is correct.

C. Acceleration of body is uniform The acceleration is $a(x) = x - 20$. Since $x$ changes with time, the acceleration is not constant and therefore not uniform. This statement is incorrect.

D. Acceleration-position graph is parabola The acceleration-position graph is given by $a(x) = x - 20$. This is a linear equation, representing a straight line, not a parabola. This statement is incorrect.