Question

A car accelerates uniformly from rest to a speed of 40.0 mi/h in 12.0 s. find (a) the distance the car travels during this time and (b) the constant acceleration of the car.

Answer 1

Xf=?
Xi=0m
Vi=0m/s
Vf=40m/s
a=?
t=12s

Xf=Xi+1/2(Vi+Vf)t
Xf= 0 +1/2 (0+40)12
Xf=1/2 (40)12
Xf=240m

Vf=Vi+at
40=0+a (12)
40=12a
a=3.33m/s^2

Answer 2

Answer:

a) 107.279m

b)$1.49m/s^2$

Explanation:

We use the second and third equations of a uniformly accelerated motion as follows;

$s=ut+\frac{1}{2}at^2...................(1)\\v^2=u^2+2as...................(2)$

where u is the initial velocity, v is final velocity, a is acceleration, t is time taken and s is distance covered.

Given;

u = 0m/s (since the car starts from rest)

v =40mi/h

t = 12s

a = ?

s = ?

substituting into equation (1), we obtain the following;

$s=0*t+\frac{1}{2}*a*12^2\\s=\frac{1}{2}*a*144\\s=72a.................(3)$

For the purpose of consistency of units, we have to first of all convert the 40mi/h to m/s as follows;

1 mile = 1609.34m

therefore;

40miles = 40 x 1609.34m = 64373.8m

Also,

1 hour = 60 x 60s =3600s.

Hence;

$40mi/h=\frac{64373.8}{3600}m/s\\40mi/h=17.88m/s$

We then make sustitution into equation (2) as follows;

$17.88^2=0^2+2*a*s\\319.69=2as....................(4)$

We then proceed to solve for s and a by solving equations (3) and (4) simultaneously.

(a) To find  s, the distance covered;

from equation (3), $a=\frac{s}{72}$. We then put this into (4) to obtain the following;

$319.69=2*\frac{s}{72}*s\\319.69=\frac{2s^2}{72}\\hence\\2s^2=319.69*72\\2s^2=23017.68\\s^2=\frac{23017.68}{2}\\s^2=11508.84\\s=\sqrt{11508.84}\\\\s=107.279m$

b) As previously stated, $a=\frac{s}{72}$;

hence,

$a=\frac{107.279}{72}\\a=1.49m/s^2$