An initially motionless test car is accelerated uniformly to 145 km/h in 8.33 seconds before striking a simulated deer. the car is in contact with the faux fawn for 0.515 seconds, after which the car is measured to be traveling at 76.5 km/h. what is the magnitude of the acceleration of the car before the collision? number m/ s what is the magnitude of the average acceleration of the car during the collision? number m/ s what is the magnitude of the average acceleration of the car during the entire test, from when the car first begins moving until the collision is over?

 before collision a = 4.90 m/s², during collision   a = -36.95 m/s² and  average a3 = 2.21 m/s²

Explanation:

Before starting the problem let's reduce all the magnitude to SI units

          V1 = 145 km / h (1000m / 1 km) (1h / 3600s) = 40.28 m / s

          V2 = 76.5 km / h = 21.25 m / s

a) In this part we can use the kinematic equations

           Vf = Vo + a t

How the vehicle starts from rest (Vo = 0)

           a = Vf / t

           a = 40.8 / 8.33

           a = 4.90 m/s²

b) During the collision we must use the relationships and momentum and amount of movement

           I = Δp

           F t = m (vf-vo)

           F = m (vf-Vo) / t

           F = m (21.25 -40.28) /0.515

           F = m (-36.95)

Now let's use Newton's second law, to look for the average acceleration

           F = ma

           a = F / m

           a = 1/m   (-36.95)m

           a = -36.95 m / s²

The negative sign indicates that acceleration opposes speed

c) The magnitude of the average acceleration, to calculate this the definition of average acceleration is used, which is the change in speed between the time the change lasts

The time that the change lasts is from when the speed is zero, until reaching the maximum speed and during the crash, bone is the total time of the movement

            t3 = 8.33 + 0.515

            t3 = 8,845 s

            ΔV = 40.8 - 21.25

            ΔV = 19.55 m / s

            a3 = ΔV / t3

            a3 = 19.55 / 8.845

            a3 = 2.21 m/s²

trig 15.3 cos

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j