57.8k views
4 votes
A 1000 kg car experiences a net force of 9500 n while decelerating from 30. 0 m/s to 23. 4 m/s. How far does it travel while slowing down?.

User Picomancer
by
7.2k points

1 Answer

3 votes

Answer:

-30 m is the answer

Step-by-step explanation:

Using the formula of newtons second law

F=ma

F=force=9500N=9500kg.m/s^

m=mass=1000 kg

a=acceleration=?m/s^

a=F/m

a=(9500kg.m/s^)/(1000kg)= 9.5m/s^

Now we can easily solve it by kinematic equation

v^=u^+2as

v = final velocity = 15.9 m/s

u = initial velocity = 30 m/s

a = acceleration = 9.5 m/s²

s = displacement = ? m

s = (v² - u²)/2a

s = ((15.9 m/s)² - (30 m/s)²)/(2 × 9.5 m/s²) = -34.06 m = -30 m (rounded to one significant figure)

In this, Negative displacement means the car is slowing down.

User Shihabudheen K M
by
8.7k points