\"A uniformly accelerated car passes three equally spaced traffic signs. The sig

Question

"A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car passes the first sign at t = 1.3 s, the second sign at t = 3.2 s, and the third sign at t = 5.1 s.

(a) What is the magnitude of the average velocity of the car during the time that it is moving between the first two signs?
ANS:13.1578 m/s

(b) What is the magnitude of the average velocity of the car during the time that it is moving between the second and third signs?
ANS:13.1578 m/s

(c) What is the magnitude of the acceleration of the car?

Explanation / Answer

a. avg velocity of car during the time it is moving between first two signs = distance/time = 25/(3.2-1.3)= 13.1578 m/s b. avg velocity of car during the time it is moving between 2nd & 3rd signs = distance/ time= 25/(5.1-3.2)=13.1578 m/s c. Let speed at 1st post be u so for 1st & 2nd post => 25= u*1.9+0.5*a*1.9^2 => 25= 1.9u+1.805a for 1st & 3rd post => 50 = u*3.8+0.5*a*3.8^2 => 50=3.8u+7.22a Solving above 2 eq a=0 m/sec ^2

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