A ball is thrown directly downward with an initial speed of 8.95 m/s from a heig
ID: 1886243 • Letter: A
Question
A ball is thrown directly downward with an initial speed of 8.95 m/s from a height of 30.2 m. After what time interval does it strike the ground?
s
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.2 m/s2 while Kathy maintains an acceleration of 5.17 m/s2.
(a) Find the time at which Kathy overtakes Stan.
s from the time Kathy started driving
(b) Find the distance she travels before she catches him.
m
(c) Find the speeds of both cars at the instant she overtakes him.
Explanation / Answer
1.
Using equation
H = U*t + 0.5*a*t^2
H = 30.2 m
U = initial speed = 8.95 m/sec
a = acceleration = 9.8 m/sec^2
So,
30.2 = 8.95*t + 0.5*9.8*t^2
4.9*t^2 + 8.95*t - 30.2 = 0
Solving above quadratic equation
t = [-8.95 +/- sqrt (8.95^2 + 4*4.9*30.2)]/(2*4.9)
t = 1.73 sec
2A.
Suppose time taken by Kathy to catch Stan is 't', then travelling time of Stan = t + 1
Now when Kathy catches Stan distance traveled by both would be same, So
Dk = Ds
Uk*t + 0.5*ak*t^2 = Us*(t + 1) + 0.5*as*(t + 1)^2
Uk = Us = initial speed of both = 0 m/sec
ak = acceleration of Kathy = 5.17 m/sec^2
as = acceleration of Stan = 3.2 m/sec^2
So,
0*t + 0.5*5.17*t^2 = 0*(t + 1) + 0.5*3.2*(t + 1)^2
5.17*t^2 = 3.2*(t + 1)^2
(t + 1)/t = sqrt (5.17/3.2) = 1.27
t + 1 = 1.27*t
t*(1.27 - 1) = 1
t = 1/0.27
t = 3.70 sec
2B.
Distance traveled by Kathy in this time will be
Dk = Uk*t + 0.5*ak*t^2
Dk = 0*3.70 + 0.5*5.17*3.70^2
Dk = 35.39 m
2C.
Vk = Uk + ak*t
Vk = 0 + 5.17*3.70
Vk = 19.13 m/sec = speed of Kathy during overtake
Vs = Us + as*(t + 1)
Vs = 0 + 3.2*(3.70 + 1)
Vs = 15.04 m/sec = speed of Stan during overtake
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