A 1000 kg car experiences a net force of 9500 N while slowing down from 30 m/s to 16 m/s. How far does it travel while slowing down?

The distance is 33.89 m

The first step is to calculate the acceleration

F= ma

force= 9500N

mass= 1000 kg

9500= 1000 × a

a= 9500/1000

= 9.5 m/s

v²= u² + 2as

30²= 16² + 2(9.5)(s)

900= 256 + 19s

900-256= 19s

644= 19s

s= 644/19

s= 33.89 m

Hence the distance traveled by the car is 33.89 m

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33.89 m

Explanation:

We must first obtain the acceleration of the car from;

F=ma

Where

F= force= 9500 N

m= mass of the car= 1000kg

a= acceleration

a= F/m= 9500/1000

a= 9.5 m/s^2

From;

V^2=u^2 + 2as

Where;

V= final velocity

u= initial velocity

s= distance covered

a= acceleration

s= v^2 -u^2/2a

s= (30)^2 -(16)^2/2×9.5

s= 900 - 256/19

s= 644/19

s= 33.89 m

friction

explanation:

they us to move forward and backwards