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Answer 1 · 2017-10-23T02:41:50

The car needs to stop in #101-2=99# #m#.

The time taken is #11.1# #s#.

Summarise the data:

#u=17.8# #ms^-1#
#v=0# #ms^-1#
#s=99# #m#
#t=?# #s#

Use #v^2=u^2+2as# to find the acceleration required.

Rearrange to make the acceleration the subject:

#a=(v^2-u^2)/(2s)=(0^2-17.8^2)/(2xx99)=-1.6# #ms^-2#

The minus sign just tells us that the acceleration was in the opposite direction to the initial velocity.

Now use #v=u+at# to find the time. Rearrange to make the time the subject:

#t=(v-u)/a=(0-17.8)/(-1.6)=11.1# #s#