An automobile and a truck start from rest and at the same instant, with the automobile initially at some distance behind the truck. The truck has a constant acceleration of 2.10 m/s2, and the automobile an acceleration of 3.4 m/s2. The automobile overtakes the truck after the truck has moved 50.0 m. (a) How much time does it take the automobile to overtake the truck? (b) How far was the automobile behind the truck initially? (c) Draw the figure of the problem
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Hi!
let y meters represent distance between truck and automobile
y+50.0 m = Distance travelled by automobile---> (1)
Now let t represent time for truck to travel 50 m
therefore
50 m = v0t + 1/2 at^2
50 = 0+ 1/2 *2.1 t^2
1.05 t^2 = 50
t = +/- √ 50/1.05
t = + 6.90 seconds
Since - time is physically impossible.
b) Now refer back to equation one
y+50.0 m = Distance travelled by automobile---> (1)
Distance travelled by automobile can be represented as v0t+1/2 at^2
y + 50.0 = 0+1/2 *3.4 t^2
y = 1.7 t^2 -50
Substituting time we got earlier
y = 1.7(6.9)^2 -50
y = 30.94 meters.
Therefore automobile was 30.9 meters behind the truck.
c)
Draw two straight lines with y being the distance between automobile and truck. Mark 50 m when they are at same length and show that the automobiles position is greater than the truck after the 50 m.
Automobile
<-------------------------------------…
Truck
<----------------------------------50 m------->
Cannot draw accurately. Hope fully you get the jist of it. A graph would be more appropriate where you can show they intersect at 50 m and then the slope of automobile is steeper than the truck.
Hope it helps!
let y meters represent distance between truck and automobile
y+50.0 m = Distance travelled by automobile---> (1)
Now let t represent time for truck to travel 50 m
therefore
50 m = v0t + 1/2 at^2
50 = 0+ 1/2 *2.1 t^2
1.05 t^2 = 50
t = +/- √ 50/1.05
t = + 6.90 seconds
Since - time is physically impossible.
b) Now refer back to equation one
y+50.0 m = Distance travelled by automobile---> (1)
Distance travelled by automobile can be represented as v0t+1/2 at^2
y + 50.0 = 0+1/2 *3.4 t^2
y = 1.7 t^2 -50
Substituting time we got earlier
y = 1.7(6.9)^2 -50
y = 30.94 meters.
Therefore automobile was 30.9 meters behind the truck.
c)
Draw two straight lines with y being the distance between automobile and truck. Mark 50 m when they are at same length and show that the automobiles position is greater than the truck after the 50 m.
Automobile
<-------------------------------------…
Truck
<----------------------------------50 m------->
Cannot draw accurately. Hope fully you get the jist of it. A graph would be more appropriate where you can show they intersect at 50 m and then the slope of automobile is steeper than the truck.
Hope it helps!
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(a) Let the automobile is h meter behind the truck,
=>For truck:-
=>By s = ut + 1/2at^2
=>50 = 0 + 1/2 x 2.10 x t^2
=>t = √47.62
=>t = 6.9 sec
(b) By s = ut + 1/2at^2
=>50 + h = 0 + 1/2 x 3.4 x (6.9)^2
=>h = 30.95 m
(c) Now you can draw the figure by taking time on one axis and the distance on the second.Initially show the automobile at the origin.
=>For truck:-
=>By s = ut + 1/2at^2
=>50 = 0 + 1/2 x 2.10 x t^2
=>t = √47.62
=>t = 6.9 sec
(b) By s = ut + 1/2at^2
=>50 + h = 0 + 1/2 x 3.4 x (6.9)^2
=>h = 30.95 m
(c) Now you can draw the figure by taking time on one axis and the distance on the second.Initially show the automobile at the origin.
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Because of writing equations here, I have uploaded the detailed solution at the link:
https://sites.google.com/site/physicsato…
Solution29.pdf
https://sites.google.com/site/physicsato…
Solution29.pdf