velocity and accelaeration?

Question

A speeding car is traveling at a constant speed 30m/s. It passes a stoped car. The stopped car accelerates at 7m/s/s. How fast will it be going when it catches up to the speeding car?

Answer 1

klao633 got the 1st formula wrong. The 2nd term on the right side of the = should have been 1/2 of what was given.

Consider the problem from the reference system of the "speeding car". Then
d = Vo*t + (1/2)*a*t^2
describes the progress of the accelerating car. Vo = -30 m/s since originally it is dropping back at that rate. At the time t, when the accelerating car catches up, d = 0. So substitute 0 in for d and solve for t. Then use t in the formula
v = a*t.

Another way: When the accelerating car catches up, the distance traveled by the 2 cars will be the same and the time period will be the same. So the average speed of the accelerating car must be 30 m/s. Since the acceleration is uniform, and it started at rest, its speed at the moment of catching the other must be 2X the average.

Answer 2

The distance equations for both cars can be written as:
x=1/2a*t^2+v0*t+x0

with:
x= distance
a= acceleration
v0= initial velocity
x0=initial position
and t=time

Here, since the stopped car doesn't start accelerating until just after the speeding car passes by, we can let x0 be zero for both cars. This makes
x1 (the speeding car's position) = 1/2*0*t^2 + 30t + 0
x2 (the accelerating car's position) = 1/2* 7 *t^2 + 0t + 0

Setting those two equations equal to one another give us
30t=7/2 *t^2
0=7/2 t^2 - 30t
0=t (7/2 t-30)
t=0 and t=60/7 seconds

The two values for t means the cars are at the same point twice. t=0 gives us the point when the stopped car starts moving, which we already knew. The second give us the time when they meet up. Plugging this value of t into the equation for the velocity of the accelerating car gives us

v2 = a*t = 7*60/7 = 60 m/s

Answer 3

D=Distance
V=Velocity
A=Accellerate

C1=Speeding Car
C2=Stopped Car accelerating

D1=Distance for C1
D2=Distnace for C2

V1 = 30 m/s is always fixed 30 m/ s
V2 = speed on time (t)

A1=0 C1 doesn't have Acceleration
A2=7

C2 Catch C1 when D1=D2

D1=V*t
D2=(A*t^2)/2 use derivative and also can plot using graph as triangle


t 1 2 3 4 5 6 7 8 9 10
---------------------------------------------------------------------------------------------------------
D1 30 60 90 120 150 180 210 240 270 300
D2 3.5 14 31.5 56 87.5 126 171.5 224 283.5 350
V2 7 14 21 28 35 42 49 56 63 70

C2 catch C1 is somewhere between t=8 and t=9


Using math

D1=D2
V1*t = (A2*t^2)/2 get in what t D2=D1 [V1 =30 and A2=7]
30t=(7t^2)/2
7t^2=60t
7t=60
t=60/7=8.57

D2=(A2*t^2)/2
D2=7t^2/2
D2=7(60/7)^2/2

D2=7(60^2 / 7 ^2) /2
D2=(60^2/7)/2
D2=257.14

C2 catch C1 in Distance 257.14 m
2D2=V2*tc tc is time when D2=D1 why 2 times D2 because the area for D2 is only a half of V*t it is like triangle

V2=2D2/tc
V2=2(257.14)/8.57
V2=60


So the answer is C2 is catch C1 on V2= 60 m/s

Answer 4

Distance=Velocity * Time + Acceleration * Time^2

For Car 1:
D=30 m/s * Time + 0*Time
D=30 m/s * Time

For Car 2:
D= 0*Time + 7 m/s/s * Time ^2
D= 7 m/s/s*Time^2

Equate both distances
30* Time= 6* Time^2
30=6*Time
30/6=Time
5 sec=Time

Finally, insert the time into the equation:
VelocityFinal = VelocityInitial + Acceleration*Time
VelocityFinal = 0 + 7 * 5
VelocityFinal = 35 m/s

Answer 5

in time t the first car will have travelled 30t. t is taken as the time the two cars catch up. the second car will have travelled 0.5*7*t^2
so 30t=3.5T^2 or t = 30/3.5sec. so speed of secont car is a*t=7*30/3.5=60m/s

Answer 6

messenger me
fortitudinousskeptic
I can do this problem. I teach physics. But I'd rather explain it than just fork over an answer