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)
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)
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]
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)
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)
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 particle moves along a straight line OX. At a time \( t \) (in seconds) the distance \( x \) (in meters) of the particle from O is given by \( x = 40 + 12t – t^3 \). How long would the particle travel before coming to rest?
(2006)
Concept: Particle comes to rest when velocity is zero. Velocity \( v = \frac{dx}{dt} = 12 - 3t^2 \). Setting \( v=0 \) gives \( 12 - 3t^2 = 0 \), so \( t=2 \text{ s} \). Initial position at \( t=0 \) is \( x(0) = 40 \text{ m} \). Position at \( t=2 \) s is \( x(2) = 40 + 12(2) - (2)^3 = 56 \text{ m} \). Distance traveled is \( |x(2) - x(0)| = |56 - 40| = 16 \text{ m} \).
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)
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.
For a particle displacement time relation is \( t = \sqrt{x} + 3 \). Its displacement when its velocity is zero:
(1999)
Concept: Velocity is the time derivative of displacement. First, express \( x \) as a function of \( t \): from \( t = \sqrt{x} + 3 \), we get \( \sqrt{x} = t - 3 \), so \( x = (t-3)^2 \). Then find velocity \( v = \frac{dx}{dt} = 2t-6 \). Set \( v=0 \) to find when it is at rest: \( 2t-6=0 \) implies \( t=3 \text{ s} \). Substitute \( t=3 \text{ s} \) back into the displacement equation: \( x(3) = (3-3)^2 = 0 \text{ m} \).
A bus travelling the first one-third distance at a speed of 10 km/h, the next one-third at 20 km/h and at last one-third at 60 km/h. The average speed of the bus is:
(1997)
Concept: Average speed is total distance over total time. Let total distance be \( D \). The time taken for each one-third distance is \( t_1 = \frac{D/3}{10} = \frac{D}{30} \), \( t_2 = \frac{D/3}{20} = \frac{D}{60} \), \( t_3 = \frac{D/3}{60} = \frac{D}{180} \). Total time \( T = t_1+t_2+t_3 = D \left( \frac{6+3+1}{180} \right) = \frac{D}{18} \). Average speed \( = \frac{D}{T} = \frac{D}{D/18} = 18 \text{ km/h} \).