Question

A car travelling $20m/s$ enters into a unbanked curve with a radius of $25m$.
If the coefficient of friction between the tires and the road is $0.75$, what is the magnitude of the change in velocity of the car if the car is going to stay in its lane?

• A. $\Delta v=3.20m/s$
• B. $\Delta v=6.40m/s$
• C. $\Delta v=12.8m/s$
• D. $\Delta v=13.6m/s$
• E. $\Delta v=26.4m/s$

Answer 1

The correct option is B $\Delta v=6.40m/s$

Given - $v_1=20m/s,r=25m,\mu =0.75$ ,

as the car is entering into an unbanked road so total required centripetal force must be provided by frictional force only , to be in its lane ,

$m{v}_2^2/r=\mu mg$

where $v_2=$ velocity of car on curve ,

therefore $v_2^2=\mu rg$

or $v_2=\sqrt{\mu rg}$

or $v_2=\sqrt{0.75\times 25\times 9.8}=13.6m/s$

therefore change in velocity $\Delta v=v_2-v_1=13.6-20=-6.40m/s$

-ive sign means velocity is decreasing .