Two cars move in opposite directions toward each other on a level, straight, one-lane road. when both drivers begin to brake, their front bumpers are 275 meters apart. the smaller car is initially moving at 25.0 m/s and the larger is initially moving at 30.0 m/s. the magnitude of the acceleration of the larger car is 20.0% greater than the magnitude of the acceleration of the smaller. determine the magnitude of acceleration for each car such that when their front bumpers touch, they have just come to rest.

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

= Acceleration of light car

= Acceleration of heavy car

Adding the equations we get

Equating

Acceleration of smaller car 2.5 m/s²

Acceleration of larger car 3 m/s²

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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards