Acar leaves the starting line at 20 m/s heading to the right at constant speed. a second car starts from rest accelerating at 2.0 m/s^2. how fast is the second car moving when it passes the first car? a) 25 m/s b) 30 m/s c) 35 m/s d ) 40 m/s e) 50 m/s

The velocity of the second car when it passes the first car is 40 m/s

Explanation:

The position and velocity of the cars is given by the following equations:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

If the velocity is constant, then a = 0 and x = x0 + v · t

When the second car passes the first car, the position of both cars is the same:

x first car = x second car

x0 + v · t = x0 + v0 · t + 1/2 · a · t²                   (x0 = 0 and v0 = 0)

v · t = 1/2 · a · t²

2 · v /a = t

2 · 20 m/s / 2.0 m/s² = t

t = 20 s

Using the equation of velocity, we can calculate the velocity of the second car at t = 20 s

v = v0 + a · t

v = 0 m/s + 2.0 m/s² · 20 s = 40 m/s

The velocity of the second car when it passes the first car is 40 m/s

we need to know the time interval during which the force acts. impulse of force = momentum change