Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
Answer:
Al's mass is 102.92 kg
Explanation:
As there are no external forces in the horizontal direction, the horizontal net force must be zero:
As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:
where the suffix i and f means initial and final respectively.
The initial momentum will be:
But, as they are at rest, initially
So, this means:
We know that the have an combined mass of 195 kg:
.
so:
.
Now, we can use the values:
where the minus sign appears as they are moving at opposite directions
and this is the Al's mass.
Answer:
Acceleration is the change in velocity over the change in time = Δv/Δt. To do these problems, you need to find out how much the speed changed and over what period of time it changed.
Snail 1 changes from 4 cm/min to 7 cm/min in 3 minutes. Subtract the starting velocity (4 cm/min) from the ending velocity (7 cm/min) then divide by the time (3 min):
Snail 1 = (7 cm/min. - 4 cm/min)/(3 minutes) = ? (remember to put down the units)
Snail 2 changed from 7 cm/min. down to 1 cm/min. in 3 minutes
Snail 2 = (1 cm/min. - 7 cm/min.)/(3 min.) = ? (note that the acceleration is negative when you slow down)
I hope this helps you
