Calculate the force needed to bring a 790–kg car to rest from a speed of 79.0 km

Question

Calculate the force needed to bring a 790–kg car to rest from a speed of 79.0 km/h in a distance of 114 m (a fairly typical distance for a non-panic stop). I have the force = -1.67 x 103 N.

Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 1.70 m. Calculate the force exerted on the car and compare it with the force found in part (a).
Force = ?
Ratio = ?

I just need the second question answered to find the force and ratio. Not sure how to. I continue to get 11143 N and 67.1 and it is incorrect.

Explanation / Answer

(a) ans

mass m=790 kg

finial speed v=0

initial speed u=79 km/h=79*5/18=21.94 m/s

distance S=114 m

this concept belongs to laws of motion

according to newtron second law

F1=ma=m[v^2-u^2]/2s

=790*[0-21.94^2]/2*114

=-790*481.4/228

=-1.67*10^3 N

(b) ans

distance s=1.7

force F2=m[v^2-u^2]/2s

=790*[0-21.94^2]/2*1.7

=-790*481.4/2*1.7

=-111846.25 N

now the ratio of two force =F2/F1=111846.25/1670=66.97 N

therefore the ratio is =66.97 N

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.