the maximum points were 8. I got 5. Can someone help me find out what I did wron
ID: 1335200 • Letter: T
Question
the maximum points were 8. I got 5. Can someone help me find out what I did wrong and help me evaluate E.
2-1 Repr. Phys. Expl. Eval.Sum Dehnitions: uaug , aaug- At constant acceleration (for vertical use y instead of T E S T 2 Due Thursday, September 24, 2015 at 12:40 pm Motion C-Level r) 1. 8pt You are driving your car east on 8th Street at a speed of 40 mph (17.9 m/s) when you encounter a red light. You bring your car to a complete stop over a distance . v& of 23 m. What was the (constant) acceleration? (Fol low the steps below.) A) Draw a sketch of the situation with a coordinate system. Define positive and directions of all relevant quantities. B) Write the three formulas for motion at constant acceleration. C) For ALL variables in the 3 formulas state their values, if they are known, or state whether they are not known. D) Then, for each formula explain why it can or annot be used to find the answer. E) Now, find the acceleration. F) If you haven't done this yet under E), check care- fully whether the units in your equations resolve to the units of acceleration. Further evaluate your an- swer (trend, limiting cases, order of magnitude, ...) + 2a Freefall acceleration : g= 9.81 m/s2 elevate 23 m (The e este formde The hird ov mola The fist formula tt distance, d= .23 Gt is tused o s not vept es eni ar not non Rx 23 m m/sExplanation / Answer
A) You have already done.
B) v = u + a*t ---(1)
v^2 - u^2 = 2*a*s ---(2)
s = u*t + 0.5*a*t^2
here, u ---> initial velcoity
v --> final velocity
t --> time taken
s --> distance travelled in t seconds
a --> acceleration
C) u = 17.9 m/s
v = 0
s = 23 m
t = ?
a = ?
D) equation (1) can not be used to find, beacuse we dont know the time taken.
equation (2) can be used to find acceleration
equation(3) can be used to find acceleration, beacuse we dont know the time taken.
E) v^2 - u^2 = 2*a*s
==> a = (v^2 - u^2)/(2*s)
= (0^2 - 17.9^2)/(2*23)
= -6.97 m/s^2
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