Answer:
The acceleration of the car is [tex]1.432 m/s^{2}[/tex]
Explanation:
We are assuming rightward coordinate positive and all quantities are along this direction
We know,
[tex]a = \frac{dv}{dt}[/tex]
where a - acceleration, v=velocity, t=time and x=displacement
multiply by dx in both sides
[tex]adx = \frac{dv}{dt} dx[/tex]
but we know [tex]\frac{dx}{dt} = v[/tex]
Therefore,
[tex]adx = vdv[/tex]
Here we integrate both sides with proper limits
x ranges from 0 to 110 as v ranges from 29 to 34
p = 0, r = 110, q = 29, s = 34
[tex]a\int\limits^r_pdx = \int\limits^s_q {v} \, dv[/tex]
a is given as constant thus can be pulled out of the integration
[tex]a[110-0] = [\frac{34^{2}}{2}- \frac{29^{2}}{2}][/tex]
Therefore,
Accelaration of the car is [tex]1.432 m/s^{2}[/tex]
Note:
Here moving to the right doesn't mean anything significant other than the fact that all quantities are pointing in that direction. Therefore obtained acceleration is also towards the right
If you know equation of motion for constant acceleration as
[tex]v^{2} -u^{2} = 2as[/tex] you can plug in values in this equation to obtain value of a
v - final velocity
u - initial velocity
s - displacement
