"a 500kg car which was initially at rest traveled with an acceleration of 5m sec rise to power -2 its kinetic energy after 4 second" using the formular of kinetic energy

1*10^5 Joules

Explanation:

Mass of the car= 500kg.

Acceleration of the car= 5m/s^-2

Time for which car is accelerated= 4 seconds

Initial Velocity=0

By using the equation of motion:

Final velocity= initial velocity + time*acceleration;

Velocity =4*5=20m/s

Kinetic energy= 1/2 * mass* velocity^2

K.E=0.5*500*20^2=1*10^5 Joules

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s