a car acclerates from 12 m/s to 25 m/s in 6.0s. What was its acceleration? How far did it ravel in this time?t

Question

constant acceleration, HOW WOULD ONE GO ABOUT SOLVING THIS?

Answer 1

it may be your homework but so what, i'll still help ;-)

the acceleration is the change in speed per unit of time. So here the car has changed from 12 m/s, to 25 m/s, that's a change of 25-12=13m/s. And it has done so in 6 seconds, so the acceleration is 13/6 = 2.17 m/s^2 (ie meters per second, per second).

for the second bit i'm afraid you need to either remember your calculus (*), or know by heart that x, the position, is given by:
x = x0 + v0*t + (1/2)*a*t^2
where x0 is the initial position (here, 0)
v0 is the initial speed (here, 12 m/s)
a is the acceleration

plug it in, you get
x = 0 + 12*6 + (1/2)*(13/6)*6*6
which gives
x = 111 meters


(*) you start with a = dv/dt, you integrate you get v = a*t + constant, and the constant is V0, so v=v0+a*t. And since v=dx/dt, you integrate one more, and you get x=v0*t+(1/2)*a*t^2+constant, and the constant is x0, so you get x=x0+v0*t+(1/2)*a*t^2

Answer 2

Acceleration = a(t) = change in velocity/change in time

= (25 - 12)/6

= 13/6 m/s^2

Velocity = v(t) = integral of a(t)

= (13/6)*t + C1***

***C1 is a constant that's evaluated using known conditions

Set t = 0 when v = 12 m/s. This gives:

12 = (13/6)*0 + C ---------> C1 = 12

Update velocity equation: v(t) = (13/6)*t + 12

To find out how far the car traveled during its period of acceleration, we need the position function, or p(t).

x(t) = integral of v(t)

= (13/6)*(t^2/2) + 12*t + C2 (like C1, C2 is also a const)

= (13/12)*t^2 + 12t + C2

To easily solve for C2, assume that the road is the x-axis
and that when t = 0, the car is at the origin (x = 0) with a velocity of 12 m/s....

x(0) = (13/12)*(0)^2 + 12*0 + C2 = 0 -------> C2 = 0

Update position function: x(t) = (13/12)t^2 + 12t

Since we assumed the car was at x = 0 when t = 0 and v =12 m/s, all that's left to do is find the value of x when t = 6 sec.

x(6) = (13/12)*(6)^2 + 12*(6)

= 39 + 72

= 111 meters

The car travels 111 meters during this acceleration period.

Answer 3

final velocity - starting velocity in this case 25 - 12 which is 13
then u put the change in velocity (final velocity - starting velocity) over the time, in this case it would be 13/6, which is about 2.17 meters per seconds squared.

Answer 4

i think this question is easy... I don't think you cannot answer this question unless you wasn't be in the class..

25 m/s - 12 m/s = 13 m/s
13 m/s for 6 seconds
so, 2.5 m per second....