A car starts from rest and travels for #5.0# #s# with a constant acceleration of #-1.5# #m##/##s^2#. What is the final velocity of the car? How far does the car travel in this time interval?

1 Answer
Shwetank Mauria
Mar 4, 2016

Final velocity #-7.5# #m/sec# and distance covered #-18.75# #m#. Minus sign indicates that acceleration is in reverse direction.

Explanation:

If an object having initial velocity #u# is applied an acceleration of #a# for #t# time, its final velocity #v# is given by #v=u+at# and distance covered #S# is given by #S=ut+1/2at^2#.

We have #u=o# (as the object starts from rest), #t=5# sec., and acceleration #a=-1.5# #m/sec^2#.

Hence final velocity #v=-1.5xx5=-7.5# #m/sec#.

Distance covered #S=1/2xx(-1.5)xx5^2=-18.75# #m#.

Minus sign indicates that acceleration is in reverse direction.

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