A moving car possesses average velocities of `5 ms^-1, 10 ms^-1 and 15 ms^-1` in the first, second and third seconds respectively. What is the total distance covered by the car in these three seconds ?

A moving car possesses average velocities of 5 m s^(-1), 10 m s^(-1) , and 15 m s^(-1) , in the first, second, and third seconds, respecticely. What is the total destance coverd by the car in these 3 s .?

A moving car possesses average velocities of 5 m s^(-1), 10 m s^(-1) , and 15 m s^(-1) , in the first, second, and third seconds, respecticely. What is the total destance coverd by the car in these 3 s .?

A car starts from rest and moves with constant acceleration. The ratio of distance covered by the car in nth second with that covered in n seconds is:

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 ?

The speed of a car increases uniformly from zero to 10ms^-1 in 2s and then remains constant (figure) a. Find the distance travelled by the car in the first two seconds. b. Find the distance travelled by the car in the next two seconds. c. Find the total distance travelled in 4s .

The speed of a car increases uniformly from zero to 10ms^-1 in 2s and then remains constant (figure) a. Find the distance travelled by the car in the first two seconds. b. Find the distance travelled by the car in the next two seconds. c. Find the total distance travelled in 4s .

A particle has an initial velocity of 5.5 ms^-1 due east and a constant acceleration of 1 ms^-2 due west. The distance covered by the particle in sixth second of its motion is.

A car is moving with a constant acceleration and initial velocity 2 ms^(-1) on a straight path. After 10 s it attains the velocity 12 ms^(-1). Then find the distance covered by it in this 10 s.