SECTION 2 A car starts from rest, and begins accelerating at a constant rate a,.
ID: 1431272 • Letter: S
Question
SECTION 2 A car starts from rest, and begins accelerating at a constant rate a,. It accelerates at this rate for a distance of G-45·5 ·m from its starting point, and then immediately begins to decelerate at a different constant rate a2, eventually coming to rest again after traveling an additional distance of H66m from where it began decelerating. The entire trip from start to finish (starting and ending at rest) lasts for a duration of s e of H E m from where it began decelerating. The entire trip S. What is the magnitude of the initial rate of acceleration, a,? Answer: Answer: What is the magnitude of the subsequent rate of deceleration, a,?Explanation / Answer
for accelerated part of journey :
acceleration = a1
displacement = G = 45.5 m
final velocity = Vf = final velocity
initial velocity = Vi = 0 m/s
time = t1
Vf2 = Vi2 + 2 a1 G
Vf2 = 02 + 2 a1 (45.5)
Vf2 = 91 a1 eq-1
Vf = a1 t1 eq-2
G = Vi t1 + (0.5) a1 t12
91 = a1 t12
a1 = 91/t12 eq-3
For decelerated motion :
initial velocity = Vi = Vf = a1 t1
acceleration = - a2
displacement = H = 66.4 m
time = t2 = 10.5 - t1
final velocity = Vf = 0
Vf = Vi + a2 t2
0 = a1 t1 - a2 (10.5 - t1 )
a1 t1 = a2 (10.5 - t1 )
a2 = a1 t1 /(10.5 - t1 ) eq-4
Vf2 = Vi2 + 2 a1 G
02 = (a1 t1 )2 - 2 a2 (66.4)
Using eq-4
0 = (a1 t1 )2 - 132.8 a1 t1 /(10.5 - t1 )
using eq-2
0 = ((91/t12) t1 )2 - 132.8 (91/t12) t1 /(10.5 - t1 )
t1 = 4.3 sec
t2 = 10.5 - t1 = 10.5 - 4.3 = 6.2 sec
a1 = 91/t12 = 91 / (4.3)2 = 4.92 m/s2
a2 = a1 t1 /(10.5 - t1 ) = 4.92 (4.3) / (10.5 - 4.3) = 3.41 sec
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