Questions 1 to 7 from RACE #17 Physics

Question

1. Find the following integrals:

1. \( \int 3x^7 \, dx \)
2. \( \int 4 \sqrt{x} \, dx \)
3. \( \int \frac{dx}{\sqrt{x}} \)
4. \( \int (x^3 - 5x^2 + 7x - 11) \, dx \)
5. \( \int \left( \sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} \right) dx \)
6. \( \int 3x \, dx \)
7. \( \int 5x^2 \, dx \)

2. A particle moves on x-axis as per equation \( x = t^3 - 9t^2 + 15t + 2 \) m. Distance travelled by the particle between \( t=0 \) and \( t=5 \) s is:

(A) 25 m (B) 39 m (C) 23 m (D) 52 m

3. Suppose velocity of a cricket ball hit by a batsman is given by \( \vec{v} = (2\hat{i} + 6\hat{j}) + (20 - 10t) \hat{k} \) m/s. Find the time \( t \) when acceleration of the ball is perpendicular to its velocity.

(A) 1 s (B) 2 s (C) 3 s (D) 4 s

4. A particle moves along a straight line and its position as a function of time is given by \( x = t^3 - 3t^2 + 3t + 3 \). Then the particle:

(A) stops at \( t=1 \) s and reverses its direction of motion

(B) stops at \( t=1 \) s and continues further without change of direction

(C) stops at \( t=2 \) s and reverses its direction of motion

(D) stops at \( t=2 \) s and continues further without change of direction

5. A ball is thrown vertically upwards with some speed. It reaches two points A and B one after another such that heights of A and B are one fourth and three-fourth of the maximum height attained. If the total time of flight is \( T \), the maximum time taken by the ball to travel from A to B is:

(A) \( \frac{(\sqrt{3}+1)}{4} T \) (B) \( \frac{(\sqrt{3}-1)}{2} T \) (C) \( \frac{(\sqrt{3}+1)}{2} T \) (D) \( \frac{T}{\sqrt{2}} \)

6. The initial velocity of a particle is \( u \) and the acceleration is given by \( kt \), where \( k \) is a positive constant. The distance travelled in time \( t \) is:

(A) \( s = ut^2 + kt^2 \) (B) \( s = ut + \frac{kt^3}{6} \) (C) \( s = ut + \frac{kt^2}{2} \) (D) \( s = \frac{ut^2}{2} + \frac{kt^3}{6} \)

7. A motor boat of mass \( m \) moves along a lake with velocity \( V_0 \). At \( t=0 \), the engine of the boat is shut down. Magnitude of resistance force offered to the boat is equal to \( rV \) (where \( V \) is instantaneous speed). What is the total distance covered till it stops completely? [Hint: \( F(x) = m \frac{dV}{dt} = -rV \)]

(A) \( \frac{mV_0}{r} \) (B) \( \frac{3mV_0}{2r} \) (C) \( \frac{mV_0}{2r} \) (D) \( \frac{2mV_0}{r} \)

Answer

1. Integrals:

1. \( \int 3x^7 dx = 3 \times \frac{x^8}{8} = \frac{3}{8} x^8 + C \)
2. \( \int 4 \sqrt{x} dx = 4 \int x^{1/2} dx = 4 \times \frac{2}{3} x^{3/2} = \frac{8}{3} x^{3/2} + C \)
3. \( \int \frac{dx}{\sqrt{x}} = \int x^{-1/2} dx = 2 x^{1/2} + C \)
4. \( \int (x^3 - 5x^2 + 7x - 11) dx = \frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 11x + C \)
5. \( \int \left( \sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} \right) dx = \int (x^{1/3} - x^{-1/3}) dx = \frac{3}{4} x^{4/3} - \frac{3}{2} x^{2/3} + C \)
6. \( \int 3x dx = \frac{3x^2}{2} + C \)
7. \( \int 5x^2 dx = \frac{5x^3}{3} + C \)

2. Distance travelled by particle \( x = t^3 - 9t^2 + 15t + 2 \) from \( t=0 \) to \( t=5 \):

Velocity \( v = \frac{dx}{dt} = 3t^2 - 18t + 15 = 3(t^2 - 6t + 5) = 3(t-1)(t-5) \)

Velocity changes sign at \( t=1 \) and \( t=5 \). So, particle changes direction at \( t=1 \).

Calculate positions:

• \( x(0) = 0 - 0 + 0 + 2 = 2 \)
• \( x(1) = 1 - 9 + 15 + 2 = 9 \)
• \( x(5) = 125 - 225 + 75 + 2 = 77 \)

Distance travelled = \( |x(1)-x(0)| + |x(5)-x(1)| = |9-2| + |77-9| = 7 + 68 = 75 \) m.

But options do not have 75 m, re-check velocity roots:

Velocity roots: \( 3t^2 - 18t + 15 = 0 \Rightarrow t^2 - 6t + 5 = 0 \Rightarrow t = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm 4}{2} \Rightarrow t=1, 5 \)

So, correct.

Distance travelled is 7 + 68 = 75 m, but options do not have 75 m. Possibly question wants displacement?

Displacement = \( x(5) - x(0) = 77 - 2 = 75 \) m.

Check if question wants distance or displacement. Since options are smaller, maybe question wants net displacement or something else.

Alternatively, check velocity sign between 0 to 1 and 1 to 5:

At \( t=0.5 \), \( v = 3(0.5-1)(0.5-5) = 3(-0.5)(-4.5) = 6.75 > 0 \) positive

At \( t=3 \), \( v=3(3-1)(3-5) = 3(2)(-2) = -12 < 0 \) negative

So particle moves forward from 0 to 1, then backward from 1 to 5.

Distance travelled = forward distance + backward distance = \( |x(1)-x(0)| + |x(5)-x(1)| = 7 + 68 = 75 \) m.

Since options do not have 75, check if question has typo or options are for displacement.

Displacement is 75 m, none matches options.

Check again the position at t=5:

\( x(5) = 125 - 225 + 75 + 2 = 77 \)

Check calculations:

\( 5^3 = 125 \)

\( 9 \times 5^2 = 9 \times 25 = 225 \)

\( 15 \times 5 = 75 \)

Sum: 125 - 225 + 75 + 2 = (125 + 75 + 2) - 225 = 202 - 225 = -23 \)

Corrected \( x(5) = -23 \)

Distance travelled = \( |9 - 2| + |-23 - 9| = 7 + 32 = 39 \) m.

Answer: (B) 39 m

3. Velocity \( \vec{v} = (2\hat{i} + 6\hat{j}) + (20 - 10t) \hat{k} \)

Acceleration \( \vec{a} = \frac{d\vec{v}}{dt} = 0\hat{i} + 0\hat{j} - 10 \hat{k} = -10 \hat{k} \)

Acceleration is constant in \( -\hat{k} \) direction.

Acceleration is perpendicular to velocity when \( \vec{a} \cdot \vec{v} = 0 \)

Calculate dot product:

\( \vec{a} \cdot \vec{v} = (-10 \hat{k}) \cdot (2\hat{i} + 6\hat{j} + (20 - 10t) \hat{k}) = -10 (20 - 10t) = -200 + 100t \)

Set to zero:

\( -200 + 100t = 0 \Rightarrow t = 2 \) s

Answer: (B) 2 s

4. Position \( x = t^3 - 3t^2 + 3t + 3 \)

Velocity \( v = \frac{dx}{dt} = 3t^2 - 6t + 3 = 3(t^2 - 2t + 1) = 3(t-1)^2 \)

Velocity is zero at \( t=1 \) only.

Since velocity is \( 3(t-1)^2 \), it is always \( \geq 0 \) and zero only at \( t=1 \).

Velocity does not change sign, so particle stops at \( t=1 \) but does not reverse direction.

Answer: (B) stops at \( t=1 \) s and continues further without change of direction.

5. Ball thrown vertically upwards, heights A and B are \( \frac{1}{4} \) and \( \frac{3}{4} \) of max height.

Let max height be \( H \), total time of flight \( T \).

Height at time \( t \): \( y = u t - \frac{1}{2} g t^2 \)

Time to reach max height \( t_m = \frac{u}{g} = \frac{T}{2} \)

Heights A and B correspond to times \( t_A \) and \( t_B \) such that:

\( y(t_A) = \frac{H}{4} \), \( y(t_B) = \frac{3H}{4} \)

Using symmetry and quadratic equation, the maximum time taken between A and B is:

\( t_{max} = \frac{(\sqrt{3} + 1)}{4} T \)

Answer: (A)

6. Initial velocity \( u \), acceleration \( a = kt \)

Velocity \( v = u + \int a dt = u + \int kt dt = u + \frac{kt^2}{2} \)

Distance \( s = \int v dt = \int \left(u + \frac{kt^2}{2} \right) dt = ut + \frac{k t^3}{6} \)

Answer: (B) \( s = ut + \frac{kt^3}{6} \)

7. Motor boat with mass \( m \), initial velocity \( V_0 \), resistance force \( F = -rV \)

Equation of motion:

\( m \frac{dV}{dt} = -r V \Rightarrow \frac{dV}{dt} = -\frac{r}{m} V \)

Separate variables:

\( \frac{dV}{V} = -\frac{r}{m} dt \)

Integrate from \( V_0 \) to 0 and 0 to \( t \):

\( \ln V \Big|_{V_0}^0 = -\frac{r}{m} t \) but \( \ln 0 \to -\infty \), so velocity approaches zero asymptotically.

Distance covered:

\( dx = V dt \)

From \( \frac{dV}{dt} = -\frac{r}{m} V \), we get \( dt = -\frac{m}{r} \frac{dV}{V} \)

So,

\( dx = V dt = V \times \left(-\frac{m}{r} \frac{dV}{V} \right) = -\frac{m}{r} dV \)

Total distance:

\( s = \int_0^{s} dx = -\frac{m}{r} \int_{V_0}^0 dV = \frac{m}{r} V_0 \)

Answer: (A) \( \frac{m V_0}{r} \)