The metre is defined as the distance travelled by ligh in `(1)/(299,792,458)` second. Why didn't people choose some easier number such as `(1)/(300.000.000)` second?

An insect starts from a point and travels in a straight path 1 mm in the first second and half of the distance covered in the previous second in the succeeding second. In how much time would it reach a point 3 mm away from its starting point.

An electron moving with the speed 5 xx 10^6 m per second is shooted parallel to the electric field of intensity 1 xx 10^3 N/C. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant, (mass of e = 9.1 xx 10^(-31) kg)

If a particle travels the first half distance with speed v_(1) and second half distance with speed v_(2) . Find its average speed during journey.

An object is moving with constant acceleration. Its velocity is 48 ms^(-1) at the end of 10 second and becomes 68 ms^(-1) at the end of 15 second. What would be the distance travelled by the object in 15 second ?

An object is moving on a linear path in definite direction with initial velocity 'u' with constant acceleration. Prove that the distance travelled by it during 'n'th second is u + (a)/(2) (2n -1).

The displacement y of a wave travelling in the x-direction is given by y = 10^(-4) sin ((600t - 2x + (pi)/(3))meters where x is expressed in meters and t in seconds. The speed of the wave-motion, in ms^(-1) , is

Two motor cars start from A simultaneously & reach B after 2 hour. The first car travelled half the distance at a speed of v_(1)=30 km hr^(-1) & the other half at a speed of v_(2)=60 km hr^(-1) . The second car covered the entire with a constant acceleration. At what instant of time, were the speeds of both the vehicles same? Will one of them overtake the other enroute?

A string vibrate according to the equation y = 5 sin ((pi x)/(3)) cos (40 pi t) where x and y in cm's and t is in second. (i) What is the equation of incident and reflected wave ? (ii) What is the distance between the adjacent nodes ? (iii) What is the velocity of the particle of the string at the position x = 1.5 cm when t = (9)/(8) sec ?

Consider the system shown in figure. We know the masses of bodies : m = 1 kg, M = 3 kg . The distance AB = BC = H//2 are covered in equal time intervals by the body of mass m and at B it undergoes an elastic collision with a stationary body of mass m_(0) (unknown). Time after this collision when the string in pu,,ey 2 becomes tight again is t_(0). t_(0) is second. t_(0)^(2) is equal to 1//x . Value of (x)/(10) is. (Given : H = 1m )

A string of length 1 m fixed one end and on the other end a block of mass M = 4 kg is suspended. The string is set into vibration and represented by equation y =6 sin ((pi x)/(10)) cos (100 pi t) where x and y are in cm and t is in seconds. (i) Find the number of loops formed in the string. (ii) Find the maximum displacement of a point at x = 5//3 cm (iii) Calculate the maximum kinetic energy of the string. (iv) Write down the equations of the component waves whose superposition given the wave. .