Acar travels from rest and accelerate uniformly for 5seconds untill it attains a velocity of 30seconds it then travels with uniform velocity for 15seconds before decelerating uniformly to rest i 10seconds sketch a graph of the motion . using the graph above calculate the acceleration during the first 5seconds . deceleration during the last 10seconds .. total distance covered throughout the motion

Acceleration during first 5 seconds is 6 m/s².

Deceleration during last 10 seconds is 3 m/s².

Total distance covered throughout the motion is 675 m.

Explanation:

The sketch showing the variation of velocity with time is given below.

Slope of the line OA gives the acceleration during first 5 seconds.

∴ Slope of line OA  =

Therefore, acceleration during first 5 seconds is 6 m/s².

Now, absolute value of slope of line BC is the deceleration during last 10 seconds.

∴ Slope of line BC  =

Therefore, deceleration during last 10 seconds is 3 m/s².

Now, area under the graph gives total distance covered.

Area under the graph is given as the sum of areas of triangle OAE, triangle BCD and rectangle BCDE.

∴Area under the graph is,

.

Hence, total distance covered is 675 m.

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )