Aracing car traveling with constant acceleration increases its speed from 10m/s to 50m/s over a distance of 60 m. how long does this take

2 seconds

Explanation:

We can solve the problem by using the following SUVAT equation:

where

S = 60 m is the distance covered

u = 10 m/s is the initial velocity

v = 50 m/s is the final velocity

t is the time taken

Re-arranging the formula and using the numbers above, we find the time taken:

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

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm