Finding deceleration from stopping distance and crash impact

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Fiorella
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Please I need help with these two problems...the answer is in the back of the book but as much as I've tried I can't find the procedure to do them:cry:

I really appreciate your help...thank you so much beforehand!

1) The operation manual of a passenger automobile states that the stopping distance is 50 m when the brakes are fully applied at 96 km/h. What is the magnitud of decelaration? What is the stopping time?

Answer: -7.1 m/s^2

2) The front end of an automobile has been designed so that upon impact it progressively crumples by as much as 0.70 m. Suppose that the automobile crashes into a solid brick wall at 80 km/h. During the collision the passenger compartment deccelerates over a distance of 0.70 m. Assume that the decceleration is constant. What is the magnitud of the decceleration?

Answer: -350 m/s^2
 

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hey! I could help u...for problem no.1
tried this equation it may help you! let:
V^2=v^2+2ad

where:
V=final velocity
v=initial velocity
a=acceleration
d= displacement (in unit of meter)

but; first you must
convert km/hr to m/s
and v=initial velocity will be zero because
it start at rest... (automobile)
 
this is easy...
ok now it says stopping distance is 50m
Since brakes are applied, final velocity is 0. Initial Velocity is 96 km/h.

so S=50m, u= 96km/h = 26.67m/s, v = 0.

so v^2 = u^2 + 2aS

so a = v^2 - u^2/ 2s = 0 - (26.67)^2/2X50 = -711.28/100 = - 7.112 m/s2
 
Oh sorry forgot about time...since v = u + at...t = v-u/a

so t = 0 - 26.67/-7.1 = -26.67/-7.1 = 3.76 sec
 
Thank you SO much you guys :biggrin:!

Homeworkboy I have a question...what does "S" and "u" stand for?

thank you again lol