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 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 \( rV \) (\( V \) 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. \( \int 3x^7 dx = 3 \cdot \frac{x^8}{8} = \frac{3}{8} x^8 + C \)
2. \( \int 4 \sqrt{x} dx = 4 \int x^{1/2} dx = 4 \cdot \frac{x^{3/2}}{3/2} = \frac{8}{3} x^{3/2} + C \)
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. \( \int \left( x^{1/3} - x^{-1/3} \right) dx = \int x^{1/3} dx - \int x^{-1/3} dx = \frac{3}{4} x^{4/3} - \frac{3}{2} x^{2/3} + C \)
6. \( \int 3x dx = \frac{3x^2}{2} + C \)
7. \( \int 5x^2 dx = \frac{5x^3}{3} + C \)

2. Distance travelled by particle:

Given \( x = t^3 - 9t^2 + 15t + 2 \)

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 \( t=5 \). Calculate position at \( t=0,1,5 \):

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

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

Check options: None matches 75 m, so re-check calculation:

Recalculate \( x(5) \):

\( 5^3 = 125, \quad -9 \times 25 = -225, \quad 15 \times 5 = 75 \)

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. Time when acceleration is perpendicular to velocity:

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

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

Condition for perpendicularity: \( \vec{a} \cdot \vec{v} = 0 \)

Calculate dot product:

\( \vec{a} \cdot \vec{v} = (-10) \times (20 - 10t) = -10(20 - 10t) = -200 + 100t \)

Set equal to zero:

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

Answer: (B) 2 s

4. Particle stops and direction:

Given \( x = t^3 - 3t^2 + 3t + 3 \)

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 \( v = 3(t-1)^2 \) 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. Time taken by ball from A to B:

Let maximum height be \( H \). Heights of A and B are \( \frac{H}{4} \) and \( \frac{3H}{4} \).

Time of flight is \( T \).

Using projectile motion relations, time to reach height \( h \) is:

\( t = \frac{T}{2} \pm \frac{T}{2} \sqrt{1 - \frac{h}{H}} \)

For A (height \( H/4 \)):

\( t_A = \frac{T}{2} \pm \frac{T}{2} \sqrt{1 - \frac{1}{4}} = \frac{T}{2} \pm \frac{T}{2} \cdot \frac{\sqrt{3}}{2} = \frac{T}{2} \pm \frac{T \sqrt{3}}{4} \)

For B (height \( 3H/4 \)):

\( t_B = \frac{T}{2} \pm \frac{T}{2} \sqrt{1 - \frac{3}{4}} = \frac{T}{2} \pm \frac{T}{2} \cdot \frac{1}{2} = \frac{T}{2} \pm \frac{T}{4} \)

Possible times:

• \( t_A = \frac{T}{2} - \frac{T \sqrt{3}}{4} \) or \( \frac{T}{2} + \frac{T \sqrt{3}}{4} \)
• \( t_B = \frac{T}{2} - \frac{T}{4} = \frac{T}{4} \) or \( \frac{T}{2} + \frac{T}{4} = \frac{3T}{4} \)

Calculate all possible time differences \( |t_B - t_A| \):

• \( \left| \frac{T}{4} - \left( \frac{T}{2} - \frac{T \sqrt{3}}{4} \right) \right| = \left| \frac{T}{4} - \frac{T}{2} + \frac{T \sqrt{3}}{4} \right| = \left| -\frac{T}{4} + \frac{T \sqrt{3}}{4} \right| = \frac{T}{4} (\sqrt{3} - 1) \)
• \( \left| \frac{3T}{4} - \left( \frac{T}{2} - \frac{T \sqrt{3}}{4} \right) \right| = \left| \frac{3T}{4} - \frac{T}{2} + \frac{T \sqrt{3}}{4} \right| = \frac{T}{4} (1 + \sqrt{3}) \)
• \( \left| \frac{T}{4} - \left( \frac{T}{2} + \frac{T \sqrt{3}}{4} \right) \right| = \left| \frac{T}{4} - \frac{T}{2} - \frac{T \sqrt{3}}{4} \right| = \frac{T}{4} (1 + \sqrt{3}) \)
• \( \left| \frac{3T}{4} - \left( \frac{T}{2} + \frac{T \sqrt{3}}{4} \right) \right| = \left| \frac{3T}{4} - \frac{T}{2} - \frac{T \sqrt{3}}{4} \right| = \frac{T}{4} (1 - \sqrt{3}) = \text{negative, so take positive } \frac{T}{4} (\sqrt{3} - 1) \)

Maximum time is \( \frac{T}{4} (1 + \sqrt{3}) = \frac{(\sqrt{3} + 1)}{4} T \)

Answer: (A) \( \frac{\sqrt{3}+1}{4} T \)

6. Distance travelled with acceleration \( a = kt \):

Acceleration: \( a = \frac{dv}{dt} = kt \)

Integrate velocity:

\( v = u + \int_0^t kt' dt' = u + \frac{k t^2}{2} \)

Distance:

\( s = \int_0^t v dt = \int_0^t \left( u + \frac{k t'^2}{2} \right) dt' = ut + \frac{k}{2} \cdot \frac{t^3}{3} = ut + \frac{k t^3}{6} \)

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

7. Distance covered by motor boat:

Given \( F = m \frac{dV}{dt} = -rV \)

Rewrite as:

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

Separate variables:

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

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

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

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