An object with a mass of 1.5 kg changes its velocity from +15 m/s to +22 m/s during a time interval of 3.5 seconds. what impulse was delivered to the object?

The answer is 10.5 kg m/s

Impulse (I) is the multiplication of force (F) and time interval (Δt): I = F · Δt

Force (F) is the multiplication of mass (m) and acceleration (a): F = m · a

Acceleration (a) can be expressed as change in velocity (v) divided by time interval (Δt): a = Δv/Δt

So: 
a = Δv/Δt         ⇒ F = m · a = m · Δv/Δt
F = m · Δv/Δt   ⇒ I = m · Δv/Δt · Δt
Since Δt can be cancelled out, impulse can be expressed as:
I = m · Δv = m · (v2 - v1)

It is given:
m = 1.5 kg
v1 = 15 m/s
v2 = 22 m/s

I = 1.5 · (22 - 15) = 1.5 · 7 = 10.5 kgm/s.

The answer is: 11N·s

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.