a student is taking a car for a joy ride on a long straight highway. at the begi
ID: 1688663 • Letter: A
Question
a student is taking a car for a joy ride on a long straight highway. at the beginning of the joy ride, the student has the car accelerate from rest to some top speed with a constant acceleration of 4.5 m/s^2 and then maintains that top speed to the end of the ride. the total tome taken from the start of the joy ride to its end is 1:36 mins an the total distance covered is 3.84 km. what was the time elapsed during the acceleration phase of the cars motion? what distance did the car cover during the acceleration phase of its motion? what was the top speed reached by the car?Explanation / Answer
Given Initial velocity, vi = 0 m/s car is moving with constant acceleration, a = 4.5 m/s^2 The distanc travelled during this phase is, s1 = 1/2at1^2 In the next phase, the car is moving with constant speed, then the distance travelled in this phase is,s2 = vt2 Total time taken for the entire ride, T = 1:36 min =5760 s Total distance covered,S = 3.84 km ------------------------------------------------------------------------------------- Total distance travelled = s1+s2 3.84 *1000 m = 1/2at1^2+ vt2 [since v= at1] 3840 = 1/2at1^2+at1t2 3840 = at1(t1/2+t2)-------(1) Since T = t1 +t2 5760 = t1+ t2 t2= 5760-t1 substitute this value in (1) 3840 = at1(t1/2+(5760-t1)) on solving, we get t1^2-11520t1+1706.6 =0 t1 =0.14 s t2 = 5760-0.14 =5759.86 s ------------------------------------------------------------- The time elapsed during the acceleration phase is 0.14 s The distance travelled in this phase is s1 = 1/2at1^2 = (0.5)(4.5 m/s^2)(0.14)^2 = 0.04 m s2 =3840 -s1 =3840-0.04 = 3839.96 m ----------------------------------------------------------------- The top speed acquired by the car is s2 = vt2 3839.96 = v (5759.86) v = 0.66 m/sRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.