Step 1: Identify the given parameters.
The mass, initial conditions, and force acting on the body are given.
m = 1.0 kg m = 1.0\,\text{kg} m = 1.0 kg u ⃗ = 0 at t = 0 \vec{u} = 0 \quad \text{at} \quad t = 0 u = 0 at t = 0 F ⃗ = α t i ^ + β j ^ \vec{F} = \alpha t\,\hat{i} + \beta\,\hat{j} F = α t i ^ + β j ^ α = 1.0 Ns − 1 , β = 1.0 N \alpha = 1.0\,\text{Ns}^{-1}, \quad \beta = 1.0\,\text{N} α = 1.0 Ns − 1 , β = 1.0 N Step 2: Determine the acceleration of the body.
Apply Newton's second law, a ⃗ = F ⃗ m \vec{a} = \frac{\vec{F}}{m} a = m F
a ⃗ = α t m i ^ + β m j ^ \vec{a} = \frac{\alpha t}{m}\,\hat{i} + \frac{\beta}{m}\,\hat{j} a = m α t i ^ + m β j ^ a ⃗ = t i ^ + j ^ \vec{a} = t\,\hat{i} + \hat{j} a = t i ^ + j ^ Step 3: Calculate the velocity of the body.
Integrate the acceleration with respect to time.
v ⃗ ( t ) = ∫ a ⃗ d t \vec{v}(t) = \int \vec{a}\,dt v ( t ) = ∫ a d t v ⃗ ( t ) = t 2 2 i ^ + t j ^ \vec{v}(t) = \frac{t^2}{2}\,\hat{i} + t\,\hat{j} v ( t ) = 2 t 2 i ^ + t j ^ At t = 1.0 s t = 1.0\,\text{s} t = 1.0 s
v ⃗ ( 1.0 ) = 1 2 ( i ^ + 2 j ^ ) m/s \vec{v}(1.0) = \frac{1}{2}(\hat{i} + 2\hat{j})\,\text{m/s} v ( 1.0 ) = 2 1 ( i ^ + 2 j ^ ) m/s Thus, statement (3) is true.
Step 4: Calculate the position of the body.
Integrate the velocity with respect to time.
r ⃗ ( t ) = ∫ v ⃗ d t \vec{r}(t) = \int \vec{v}\,dt r ( t ) = ∫ v d t r ⃗ ( t ) = t 3 6 i ^ + t 2 2 j ^ \vec{r}(t) = \frac{t^3}{6}\,\hat{i} + \frac{t^2}{2}\,\hat{j} r ( t ) = 6 t 3 i ^ + 2 t 2 j ^ At t = 1.0 s t = 1.0\,\text{s} t = 1.0 s
r ⃗ ( 1.0 ) = 1 6 i ^ + 1 2 j ^ \vec{r}(1.0) = \frac{1}{6}\,\hat{i} + \frac{1}{2}\,\hat{j} r ( 1.0 ) = 6 1 i ^ + 2 1 j ^ The magnitude of the position vector is:
∣ r ⃗ ( 1.0 ) ∣ = ( 1 6 ) 2 + ( 1 2 ) 2 = 10 6 ≈ 0.53 m |\vec{r}(1.0)| = \sqrt{\left(\frac{1}{6}\right)^2 + \left(\frac{1}{2}\right)^2} = \frac{\sqrt{10}}{6} \approx 0.53\,\text{m} ∣ r ( 1.0 ) ∣ = ( 6 1 ) 2 + ( 2 1 ) 2 = 6 10 ≈ 0.53 m Thus, statement (4) is false.
Step 5: Calculate the torque acting on the body.
The torque about the origin is defined as τ ⃗ = r ⃗ × F ⃗ \vec{\tau} = \vec{r} \times \vec{F} τ = r × F
τ ⃗ = ( t 3 6 i ^ + t 2 2 j ^ ) × ( t i ^ + j ^ ) \vec{\tau} = \left(\frac{t^3}{6}\,\hat{i} + \frac{t^2}{2}\,\hat{j}\right) \times (t\,\hat{i} + \hat{j}) τ = ( 6 t 3 i ^ + 2 t 2 j ^ ) × ( t i ^ + j ^ ) τ ⃗ = ( t 3 6 − t 3 2 ) k ^ \vec{\tau} = \left(\frac{t^3}{6} - \frac{t^3}{2}\right)\,\hat{k} τ = ( 6 t 3 − 2 t 3 ) k ^ τ ⃗ = − t 3 3 k ^ \vec{\tau} = -\frac{t^3}{3}\,\hat{k} τ = − 3 t 3 k ^ At t = 1.0 s t = 1.0\,\text{s} t = 1.0 s
τ ⃗ ( 1.0 ) = − 1 3 k ^ N ⋅ m \vec{\tau}(1.0) = -\frac{1}{3}\,\hat{k}\,\text{N}\cdot\text{m} τ ( 1.0 ) = − 3 1 k ^ N ⋅ m The magnitude of the torque is:
∣ τ ⃗ ∣ = 1 3 N ⋅ m |\vec{\tau}| = \frac{1}{3}\,\text{N}\cdot\text{m} ∣ τ ∣ = 3 1 N ⋅ m Thus, statement (1) is true and statement (2) is false.
Step 6: Identify the correct statements.
Based on the calculations, statements (1) and (3) are true.
Answer: Statements (1) and (3) are true.
