At the bottom of a hill, a car has an initial velocity of +16.0 meters per second. The car is uniformly accelerated at -2.20 meters per second squared for 5.00 seconds as it moves up the hill. How far does the car travel during this 5.00-second interval?
1.) 107m
2.) 74.5m
3.) 52.5m
4.)25.0m

In order to get the distance traveled by car, we will use the equation of motion. The equation of motion for calculating the distance is expressed as:

[tex]S = ut+\frac{1}{2}at^2[/tex] where;

u is the initial velocity
t is the time taken
a is the acceleration

Given the following parameters

u = 16m/s

a = -2.20m/s²

t = 5secs

Substitute the given values into the expression as shown:

[tex]S = 16(5)+\frac{1}{2}(-2.20)(5)^2\\S=80-1.10(25)\\S=80-27.5\\S= 52.5m[/tex]

Hence the car travels for 52.5m during the 5 seconds interval.

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At the bottom of a hill, a car has an initial velocity of +16.0 meters per second. The car is uniformly accelerated at -2.20 meters per second squared for 5.00 seconds as it moves up the hill. How far does the car travel during this 5.00-second interval? 1.) 107m 2.) 74.5m 3.) 52.5m 4.)25.0m

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At the bottom of a hill, a car has an initial velocity of +16.0 meters per second. The car is uniformly accelerated at -2.20 meters per second squared for 5.00 seconds as it moves up the hill. How far does the car travel during this 5.00-second interval?
1.) 107m
2.) 74.5m
3.) 52.5m
4.)25.0m