5. A soapbox derby race car starts at rest and moves 300 m down a track in 22.4 s. Calculate its acceleration (assumed constant) and its speed at the end of the track.
6. A car traveling at 35 mi/hr is stopped in a distance of 2.5 ft when it hits a tree. (a) Calculate the average deceleration of the car in m/s2. (b) Calculate the time needed to stop the car (assuming constant deceleration).
7. An automobile and a truck start from rest at the same instant, with the automobile initially at some distance behind the truck. The truck has a constant acceleration of 2 m/s2 and the automobile an acceleration of 3 m/s2. The automobile overtakes the truck after the truck has moved 75 m.
a) How long does it take the automobile to overtake the truck?
b) How far was the car behind the truck initially?
c) What is the velocity of each when they are abreast?
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Expert's answer
2020-12-10T11:09:51-0500
Explanations & Calculations
5)
Since the acceleration is constant any of the 4 motion equations can be used & by using appropriately,
s300a=ut+21at2=0+21a(22.4)2=1.196ms−2 : Since s & t are known
s300v=2(v+u)t=2(v+0)×22.4=26.786ms−1 : Can be calculated using other equations as a is known by then
6)
Again as a constant deceleration, any of the 4 motion equations can be used.
35mi/hr=(3600)s35×1.609×1000m=56315ms−1
2.5ft=2.5×0.3048m=0.762m
(a) using v2=u2+2as
02a=563152+2a×0.762=−2.08×109ms−2
(b)using s=2(v+u)t
0.762t=2(0+56315)t=2.706×10−5s
7)
Both the vehicles spend the same time until they meet. Therefore, truck travels 75m during that period while the a.m travels (x+75)m. (if the initial separation between them is x)
(a)Then apply s=ut+21at2 for the motion of the truck.
75tt=0+21×2ms−2×t2=±275×2=8.66s
(b) By applying s=ut+21at2 for the automobile, for its motion until the overtake
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