How Does the Distance Traveled by the Truck Compare to That of the Car?

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The discussion centers on comparing the distances traveled by a car and a truck under different acceleration conditions. The car accelerates from rest to 20 m/s in 5 seconds, while the truck decelerates from 20 m/s to a stop over 10 seconds. Participants emphasize the need to calculate average velocities and use kinematic equations to determine the distances. Initial calculations suggest the car travels 100 meters and the truck 200 meters, though there is uncertainty about the accuracy. Understanding constant acceleration equations is crucial for resolving the comparison.
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On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a constant rate from 20 m/s to a complete stop over a 10 second interval. How does the distance traveled by the truck compare to that of the car?
 
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sarah111 said:
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a constant rate from 20 m/s to a complete stop over a 10 second interval. How does the distance traveled by the truck compare to that of the car?
Hello sarah111. Welcome to PF !

What have you tried?

Where are you stuck?


You should show some work before anyone gives help.

... but, here's a hint:
What is the average velocity of each vehicle?​
 
I feel like I can't answer the question because i don't know what the accceleration is. I tried using the equation for speed and found that the car stop 100m and the truck 200 but i don't feel that is right
 
sarah111 said:
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a constant rate from 20 m/s to a complete stop over a 10 second interval. How does the distance traveled by the truck compare to that of the car?

Sketch a velocity vs. time graph.
The car starts from rest and constantly increasing the speed until finally attained 20m/s after 5 secs.
The truck starts from 20m/s and constantly reducing the speed till zero.
Distance traveled is - velocity x time
 
sarah,

What equations do you know regarding constant acceleration?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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