A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/hr and second half of the distance at speed v. If the average speed is 48 km/hr, then v is ____ .
S = 0.2 km u1 = 40 km/hr u2 = v km/hr 0.2/48 = 0.1/40 + 0.1/v On solving, v = 60 km/hr
A stone is thrown vertically up from the ground with an initial velocity of u. It attains a height of 80 m twice, the interval being 6 s. The initial velocity is ____ . (Assume g = 10 ms-2)
u = u ms-1, a = - g = -10 ms-2, s = 80 m We know s = ut+1/2at2 Substituting the values, 80 = ut - 5t2 Or, 5t2 -ut + 80 = 0 Or, t2-( u /5)t + 16 = 0 Solving, we get t=u+ √( u2−16000)/10and u− √( u2−1600)/ 10 It is given that the time interval is 6 s. Hence, u+ √( u2−1600)/10− (u−√( u2−1600)/10)= 6 u = ± 50 ms-1 Hence, initial velocity is 50 ms-1.
A ball is thrown up from the top of a tower 50 m high with a velocity of 15 ms-1. The time it takes to strike the ground is ____ .(Assume g = 10 ms-2)
u = +15 ms-1, a = -10 ms-2and s = -50 m We know s = ut + 1/2 at2 Substituting the values, we have -50 =15 t - 5t2 Or , t2 - 3t - 10 = 0 Hence, t can be 5 s or -2 s. Since negative value is not possible, the time that the ball takes to strike the ground is 5 s.
The displacement of a particle is given by x (t) = t2 − 2 t m. Which of the following regarding its instantaneous velocity is true?
Given by x (t) = t2 − 2 t m, Velocity at any instant is given by v (t) = dx/dt=d/dt (t2−2t) = 2t −2 = 2(t − 1) m/s Hence, velocity at t = 0 s is v (0) = 2× (0 −1) = −2 m/s Velocity at t = 1 s is v (1) = 2× (1 − 1) = 0 m/s Velocity at t = 2 s is v (2) = 2× (2 − 1) = 2 m/s Velocity at t = 3 s is v (3) = 2× (3 −1) = 4 m/s Velocity at t = 4 s is v (4) = 2× (4 − 1) = 6 m/s Velocity at t = 5 s is v (5) = 2× (5 − 1) = 8 m/s As observed, the velocity of the particle always increases with time.
The displacement of a particle is given by x(t)= 4 t2 +8 t m. Its velocity at t = 5 s is ____ m/s.
Given x (t) = 4 t2 + 8 t m Velocity at any instant is given by v(t) = dx/dt=d/dt(4t2+8t) = 8t + 8 Hence, velocity at t = 5 s is v(5) = 8 × 5 + 8 = 48 m/s
The displacement x of a particle varies with time t as x = ae- αt + beβt, where a, b, α, β are positive constants. The velocity of the particle will ____ .
x = ae−αt + beβt Differentiating x with t, gives the equation for velocity, v = −αae−αt + βbeβt Differentiating v with t, gives the equation for acceleration, Acceleration = α2ae−αt +β2beβt Since acceleration is positive, velocity would go on increasing with time.
A ball thrown up is caught by the thrower, 4 seconds after start. The height to which the ball has risen is ____.
Time of ascent (t1) = time of descent = (t2) ∴ t1 + t2 = 4 t2 = 2 seconds u = 0; a = 10 m s-2; t = 2 s; s = ? s = ut + 1/2 at2 s = 0 + 1/2×10×22 = 20 m
A train 100 m long is moving with a velocity of 60 km/hr. The time it takes to cross a bridge 1 km long is ____.
Total distance to be travelled = S = 100 + 1000 = 1100 m Velocity v = 60 km/hr = 60×5/18 = 50/3m/s Therefore, time taken t = sv t = ( 1100 )/(50/3) t = 66 seconds
The displacement of a body is given by the equation, x(t) = 3t2 + 2 t m. Its velocity at t = 2 s is ____m/s.
Given x(t) = 3/ t2 + 2 t m Velocity at any instant is given by v (t) = dx/dt=d/dt(3t2+2t) = 6t + 2 Hence, velocity at t =2 s is v(2) = 6 × 2 + 2 = 14 m/s
Wind is blowing west to east along two parallel tracks. Two trains moving with same speed in the opposite directions have the steam track of one double than other. The speed of train is ____.
Please select 2 correct answers
Let u & v be the speed of train and wind respectively. The speed of steam track of train moving in the direction of wind = u - v The speed of steam track of train moving in the opposite direction of wind = u + v. As per question, (u + v) = 2 ( u - v ) => u + v = 2u - 2v => u = 3v
A train which is 150 m long is moving due south at a 10 ms-1. A parrot flies at a speed of 5 ms-1 towards north, parallel to the track. The parrot shall cross the train in ____
Relative Velocity of parrot w. r. t. train = 5 + 10 = 15 ms-1 Time taken = 150/15 = 10 seconds
A body starts from rest and moves with uniform acceleration. The ratio of the distance covered in n seconds to the distance covered in the nth second is ____.
Distance covered in n seconds is given by s = 0×n + 1/2× a×n2 (The body starts from rest) Distance covered in the nth second is given by sn = 0 + a/2(2n-1). Hence, ratio of s/ sn = [an2/2]/ [a/2(2n-1)] = n2/2n-1
A body travels 5 km due north and 5 km due south. The distance and displacement traversed are ____ and ____ .
The distance and displacement traversed are 10 km and 0 km respectively. Distance traversed = 5 km + 5 km = 10 km Displacement = 0 km (as the body comes back to the original position)
A particle moves with a uniform velocity. The option which is TRUE about the motion of particles is ____.
The velocity of a body depends on time according to the equation v = 20 + 0.1 t2. Then, the body is undergoing ____,
Acceleration =dv/dt = 0.1×2t = 0.2 t As it depends upon time 't', acceleration is non-uniform (as time varies).
Two persons A and B are walking with speed 4 km/hr and 5 km/hr respectively in the same direction. The distance between B and after 3 hours will be ____.
Relative velocity of B w. r.t. A = VBA = VB – VA = 5 - 4 = 1 KM/HR Distance of B ahead of A in time 't' S = VAB ×t = 1 × 3 = 3 km, S = 3000 m
A racing car moving with constant acceleration covers two successive kilometres in 30 s and 20 s respectively. The acceleration of the car will be ____.
s = ut + 1/2at2 For the first kilometre, 1000 =( u×30 )+( 1/2×a×302) i.e. 100 = 3u + 45a ................(1) By the end of the second kilometre, 2000 = (u×50) + (1/2×a×502 ) i.e. 200 = 5u + 125a ....................(2) Multiplying (2) by 3 and (1) by 5 600 =15 u + 375 a....................(3) 500 = 15 u + 225 a.....................(4) Subtracting (4) from (3) 100 = 150 a a = 150/100 = 1.5 ms-2 i.e. a = 32 ms-2
A car travelling at 58 km/hr overtakes another car travelling at 40 km/hr. Assuming each car to be 5m long, the time for overtaking is ____.
Relative velocity = (58 - 40) = 18 km/hr = 18×5/18 = 5m/s Total distance travelled = 10 m (5 - 15) Time = 10/5 = 2 seconds
The graph of displacement versus time of a body is a straight line making a positive angle with the X-axis. Then the instantaneous velocity of the body at any point is:
If the graph of displacement versus time of a body is a straight line making a positive angle with the X-axis, then the body is moving with constant velocity. Thus, its instantaneous velocity at any point is equal to its average velocity.
A car travels 300 m in 10s. It covers 48 m in the 10th second. The acceleration of the car is ___ .
Note that we do not know if the car is starting from rest. Hence, using the formula, s = ut + 1/2at2, 300 = u×10 + 1/2×a×100 -----(1) We also know that sn = u + a/2( 2n-1) Hence, 48 = u + a/2(20-1) ---------------- (2) Solving the 2 equations, we get a = 4 ms-2.
An object of mass 'm' is covering a distance 'x' in proportion to t 3/2, where 't' is the time elapsed. The conclusion that can be drawn about the acceleration is that ____
x= kt3/2 (k = constant of proportionality) v=dx/dt=d(kt3/2)/dt v=3/2kt1/2 Acceleration a=dv/dt =d/dt(3/2kt1/2) a=3kt−1/2/4 =3k/4 √t ⇒ a = 3k/4√t ⇒a∝ 1/√t ∴ Acceleration decreases with time.
If the options given, ____ is an object that is said to be in motion.
By definition, a body is said to be in motion if it changes its position with respect to its surroundings with the passage of time. While the railway track is stationary, the train moves relative to it. Hence, it is said to be in motion. In the other options, the boy/ man/ lady are stationary with respect to the train/ motorcycle or bicycle respectively. However, the train, motorcycle and bicycle are in motion by themselves with respect to the surroundings.
The displacement of a particle is given by x (t) = 5t2 – 20 t m. Its instantaneous velocity is zero at t =____ s.
Velocity at any instant is given by v(t) = dx/dt=d/dt(5t2−20t) = 10t −20 = 2(5 t − 10) m/s For the instant velocity to become zero, we equate the expression for instant velocity to zero. Thus, we get 2(5 t −10) = 0 ⇒5 t − 10 = 0 5 t = 10 or t = 2 s
A body have zero velocity and still be accelerating when ____ .
When a body is projected vertically upwards, at its highest point, the velocity of the body is zero, but its acceleration is not zero. Acceleration is equal to acceleration due to gravity acting downwards.
A motor car is travelling at 30 ms-1 on a circular road of radius 500 m. It is increasing its speed at the rate of 2ms-2. The acceleration will be ____.
v = 30 ms-1; r = 500 m ; ar = 2 ms-2 Centripetal acceleration ac= v2/r = 30×30/500 = 1.8 ms-2
A constant force is acting perpendicular to the velocity of a particle. For this situation ____,
When a constant force is acting perpendicular to the velocity, the body will describe a circular path and its acceleration (called centripetal acceleration) will be constant.
The work done by centripetal force is ____.
The velocity vector of the body at any instant acts along the tangent to the circular path and is always directed towards the centre. As θ = 900, cos 900 = 0. ∴ Work done is 0.
A driver is driving a car at 36 km per hour. When he applies brakes, they cause a deceleration of 3 ms-2. If he takes 1s to apply the brakes, ____ is the distance travelled by the car during this period.
At the moment when the driver applies brakes, the car is travelling at a speed of 10 ms-1 (36 km per hour = 10 ms-1) Hence, u = 10 ms-1, v = 0, a = -3 ms-2 We know, v2 - u2 = 2as 02 - 102 = 2 × (-3) × s s = 16.67 m
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