Acar traveling 56.0 km/h is 24.0 m from a barrier when the driver slams on the brakes. the car hits the barrier 2.00 s later. (a) what is the magnitude of the car’s constant acceleration before impact? (b) how fast is the car traveling at impact?

(A) acceleration = -3.6m/s². This is deceleration or negative acceleration.

(B) Therefore, velocity just before impact, v = 8.4m/s

Explanation:

(a) Initial velocity, u = 56km/h = (56*1000/3600)m/s = 15.6m/s

Distance, s = 24.0m, time, t = 2.0s, acceleration, a = ?

Using the equation, s = ut + 1/2at²

24.0m = 15.6m/s * 2s + (a * 2²)/2

24.0 = 31.2 +2a

2a = -31.2 + 24

2a = -7.2

a = -7.2/2

acceleration = -3.6m/s². This is deceleration or negative acceleration.

(b) final velocity just before impact, v = ? initial velocity, u = 15.6m/s, acceleration = -3.6m/s², time, t = 2.0s

using the equation v = u + at

v = 15.6m/s + (-3.6m/s² * 2s)

v = 15.6m/s - 7.2m/s

v = 8.4m/s

Therefore, velocity just before impact, v = 8.4m/s

mistakes are ok

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