The displacement \( x \) of a particle varies with time \( t \) as \( x = ae^{-\alpha t} + be^{\beta t} \), where \( a, b, alpha \) and \( beta \) are positive constants. The velocity of the particle will
(2005)
1. Be independent of \( \beta \)
2. Drop to zero when \( \alpha = \beta \)
3. Go on decreasing with time
4. Go on increasing with time
View Answer
Concept: Velocity is the time derivative of displacement. Calculate \( v = \frac{dx}{dt} = -a\alpha e^{-\alpha t} + b\beta e^{\beta t} \). The term \( -a\alpha e^{-\alpha t} \) decreases in magnitude (approaching zero), while the term \( b\beta e^{\beta t} \) increases exponentially. Thus, the velocity of the particle will go on increasing with time.
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time \( t_1 \). On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time \( t_2 \). The time taken by her to walk up on the moving escalator will be:
(2017-Delhi)
1. \( \frac{t_1 t_2}{t_2 - t_1} \)
2. \( \frac{t_1 t_2}{t_2 + t_1} \)
3. \( t_2 - t_1 \)
4. \( \frac{t_1 + t_2}{2} \)
View Answer
Concept: Relative velocity.
If Preeti's speed is \( v_p \) and escalator's speed is \( v_e \), for total length \( L \), \( v_p = L/t_1 \) and \( v_e = L/t_2 \). When Preeti walks on moving escalator, effective speed is \( v_p + v_e \). Time taken \( T = L / (v_p + v_e) = L / (L/t_1 + L/t_2) = \frac{t_1 t_2}{t_1 + t_2} \).
Two cars P and Q start from a point at the same time in a straight line and their positions are represented by \( X_P(t) = at + bt^2 \) and \( X_Q(t) = ft – t^2 \). At what time do the cars have the same velocity?
(2016 – II)
1. \( \frac{a+f}{2(1+b)} \)
2. \( \frac{f-a}{2(1+b)} \)
3. \( \frac{a-f}{1+b} \)
4. \( \frac{a+f}{2(b-1)} \)
View Answer
Concept: Velocity is the time derivative of position. Calculate \( V_P(t) = \frac{dX_P}{dt} = a + 2bt \) and \( V_Q(t) = \frac{dX_Q}{dt} = f - 2t \). Equate \( V_P(t) = V_Q(t) \) to find time \( t \). \( a + 2bt = f - 2t \) ⇒ \( 2t(b+1) = f-a \), so \( t = \frac{f-a}{2(b+1)} \).
A particle covers half of its total distance with speed \( v_1 \) and the rest half distance with speed \( v_2 \). Its average speed during the complete journey is:
[2011 Mains]
1. \( \frac{v_1 v_2}{v_1 + v_2} \)
2. \( \frac{2v_1 v_2}{v_1 + v_2} \)
3. \( \frac{v_1 v_2^2}{v_1 + v_2^2} \)
4. \( \frac{v_1 v_2}{2} \)
View Answer
Concept: Average speed is total distance divided by total time. Let total distance be \( D \). Time for first half: \( t_1 = \frac{D/2}{v_1} \). Time for second half: \( t_2 = \frac{D/2}{v_2} \). Total time \( T = t_1 + t_2 = D \left( \frac{v_1 + v_2}{2v_1 v_2} \right) \). Average speed \( = \frac{D}{T} = \frac{2v_1 v_2}{v_1 + v_2} \).
A car moves from X to Y with a uniform speed \( v_u \) and returns to Y with a uniform speed \( v_d \). The average speed for this round trip is:
(2007)
1. \( \sqrt{v_u v_d} \)
2. \( \frac{v_d v_u}{v_d + v_u} \)
3. \( \frac{v_u + v_d}{2} \)
4. \( \frac{2v_d v_u}{v_d + v_u} \)
View Answer
Concept: Average speed is total distance divided by total time. Let distance from X to Y be \( D \). Time taken to go to Y: \( t_u = D/v_u \). Time taken to return to X: \( t_d = D/v_d \). Total distance \( = 2D \). Total time \( = t_u + t_d = D/v_u + D/v_d = D \frac{v_u + v_d}{v_u v_d} \). Average speed \( = \frac{2D}{D \frac{v_u + v_d}{v_u v_d}} = \frac{2v_u v_d}{v_u + v_d} \).
A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap respectively is:
(2006)
1. 0, 0
2. 0, 10 m/s
3. 10 m/s, 20 m/s
4. 20 m/s, 0
View Answer
Concept: Average velocity is total displacement over total time. For a complete circular lap, displacement is zero, so average velocity is \( 0 \). Average speed is total distance over total time. Total distance is circumference \( 2\pi R = 2 \times 3.14 \times 100 = 628 \text{ m} \). Total time is \( 62.8 \text{ s} \). Average speed \( = 628/62.8 = 10 \text{ m/s} \).
A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity \(30\text{ km/h}\) and \(40\text{ km/h}\) respectively. The velocity of the car midway between P and Q is:
(1988)
1. 33.3 km/h
2. \(20\sqrt{2}\text{ km/h}\)
3. \(25\sqrt{2}\text{ Km/h}\)
4. 35 km/h
View Answer
Concept: Equations of motion under uniform acceleration.
Formula: \(v^2 = u^2 + 2as\).
Solution: Let \(u_P=30\), \(v_Q=40\) and distance be \(s\). \(v_Q^2 = u_P^2 + 2as\) gives \(40^2 = 30^2 + 2as\) => \(1600 = 900 + 2as\) => \(2as = 700\) => \(as = 350\). For the midway point, \(v_m^2 = u_P^2 + 2a(s/2) = u_P^2 + as = 30^2 + 350 = 900 + 350 = 1250\). So, \(v_m = sqrt{1250} = 25\sqrt{2}\text{ km/h}\).