An automobile accelerates from rest at 2.3 m/s2 for 16.2 s. The speed is then held constant for 18.4 s, after which there is an acceleration of -2.5 m/s2 until the automobile stops. What is the total distance traveled

Explanation:

Given

For first part acceleration is

time

distance traveled is given by

where s=displacement

u=initial velocity

a=acceleration

t=time

For second part speed acquired is held constant for

Distance traveled during this time

For the last part it decelerate with until the car stops so

distance moved in last part is given by

here

Total distance

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

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