Physics Problem Help Please?
Here is the problem:
At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 1.8 m/s2. At the same instant a truck, traveling with a constant speed of 8.1 m/s, overtakes and passes the automobile.
(a) How far beyond the traffic signal will the automobile overtake the truck?
(b) How fast will the car be traveling at that instant?
I have no idea how to solve this. If someone can just walk me through and explain that would be great!
Update:Okay, one thing I'm not quite getting is how you got "distance = v0t + (1/2)at^2". Is that the integral?
Comments
Dтʀĸ = (8.1 m/sec) • t ... truck's displacement
... and the standard physics equation for a constantly accelerating object:
Dcaʀ = (½) • a • t² + (Vcaʀ i) • t ... initial velocity of car = Vcaʀ i = 0
Dcaʀ = (½) • (1.8) • t²
Dcaʀ = (0.9) • t²
setting the displacements equal:
(0.9) • t² = (8.1) • t
(0.9) t² − (8.1) t = 0
t • [ (0.9) t − 8.1 ] = 0
Solutions: [t = 0]
and (0.9) t − 8.1 = 0
t = 9 sec
... so the two vehicles meet at the start [t = 0] , and again at [t = 9 sec]
... and the standard physics equation for the final velocity of an accelerating object:
Vcaʀ ғ = a • t + Vcaʀ i ... final speed of car at [t = 9 sec] = Vcaʀ ғ
Vcaʀ ғ = (1.8) • (9) + 0
Vcaʀ ғ = 16.2 m/sec
... and the velocity equation is the derivative of the displacement equation.
Well, basically, distance = velocity * time, and distance = v0t + (1/2)at^2. So distance needs to be equal, i.e.:
8.1 * t = (1/2)*1.8*t^2
Solve for t, we get t = 0 (we knew that), and t = 9. 9 * 8.1 = 72.9 m (a), and speed of the automobile is 9 * 1.8 = 16.2 m/s.
Find the distance d and time t at which
d=v_truck * t
d=v0_car*t + a*t^2/2 ; since v0_car=0, d=a*t^2/2
v_truck*t = a*t^2/2 -> v_truck=a*t/2 -> t=2*v_truck/a = 2*8.1/1.8 = 9s
distance d=8.1*9 = 72.9m
speed of car: v=v0+a*t = 9*1.8=16.2m/s
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