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1 answer:
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Answer:
24.084 m/s
Explanation:
From the law of conservation of linear momentum
Total momentum before collision equals to the total momentum after collision
Since momentum=mv where m is mass and v is velocity
where
is the mass of the truck,
is velocity of the truck,
is the common velocity of moving and standing truck after collision and
is the mass of the standing truck
Making the subject we obtain
Substituting as 25000 Kg,
as 22.3 m/s,
as 2000 Kg we obtain
Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive
The truck was moving at 24.084 m/s
1) 5.79 s
2) 98.4 ft/s
Explanation:
1)
The motion of the car is a uniformly accelerated motion (it means it travels with constant acceleration), so we can find the time it takes for the car to stop by using the following suvat equation:
where
s is the distance travelled
v is the final velocity
t is the time
a is the acceleration of the car
In this problem we have:
s = 285 ft is the distance travelled
is the acceleration of the car (negative since the car is slowing down)
v = 0 ft/s is the final velocity of the car, since it comes to a stop
Solving for t, we find:
2)
The initial speed of the car can be found by using another suvat equation, namely:
where
v is the final speed
u is the initial speed
a is the acceleration
t is the time
In this problem, we have:
v = 0 is the final speed of the car
is the acceleration of the car (negative since the car is slowing down)
t = 5.79 s is the total time of motion (found in part 1)
Therefore, the initial speed of the car is: