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At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a consta...Asked by Josh
At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of 2.50 m/s^2. At the same instant, a truck, traveling with a constant speed of 15.4 m/s, overtakes and passes the automobile.
How far beyond its starting point does the automobile overtake the truck?
GOT THIS!!!! 190 m
How fast is the automobile traveling when it overtakes the truck?
Don't know this part....
v=at
a=2.5m/s^2 how long is it until v=15.4m/s?
Now s=(1/2)at^2
Use the value you found for t in this formula to find the distance.
You could also multiply it by the truck's constant speed to find how far it traveled during that time.
The auto's speed is given by the first equation above. Use the given acceleration and time you found to find v. Please post your work too.
I think I got this wrong
We have s=(1/2)at^2 and s=vt for the car and truck respectively. Thus
(1/2)at^2 = vt or 2v/a=t so your 190m looks correct when we use v=15.4m/s and a=2.50m/s^2
How far beyond its starting point does the automobile overtake the truck?
GOT THIS!!!! 190 m
How fast is the automobile traveling when it overtakes the truck?
Don't know this part....
v=at
a=2.5m/s^2 how long is it until v=15.4m/s?
Now s=(1/2)at^2
Use the value you found for t in this formula to find the distance.
You could also multiply it by the truck's constant speed to find how far it traveled during that time.
The auto's speed is given by the first equation above. Use the given acceleration and time you found to find v. Please post your work too.
I think I got this wrong
We have s=(1/2)at^2 and s=vt for the car and truck respectively. Thus
(1/2)at^2 = vt or 2v/a=t so your 190m looks correct when we use v=15.4m/s and a=2.50m/s^2
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