Explanation :
It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.
We know that acceleration is defined as the rate of change of velocity.
Where
dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s
dt is the change in time, dt = 40 s - 30 s = 10 s
So, 
From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. 
.
<span>The force would double.</span>
Answer:
6200 J
Explanation:
Momentum is conserved.
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
The car is initially stationary. The truck and car stick together after the collision, so they have the same final velocity. Therefore:
m₁ u₁ = (m₁ + m₂) v
Solving for the truck's initial velocity:
(2700 kg) u = (2700 kg + 1000 kg) (3 m/s)
u = 4.11 m/s
The change in kinetic energy is therefore:
ΔKE = ½ (m₁ + m₂) v² − ½ m₁ u²
ΔKE = ½ (2700 kg + 1000 kg) (3 m/s)² − ½ (2700 kg) (4.11 m/s)²
ΔKE = -6200 J
6200 J of kinetic energy is "lost".
