Respuesta :
Using kinematic equation s=ut + 1/2 at^2(u = initial velocity=0, s=120m, t= 6.32s), 120 = 0(t) + 1/2 a(6.32)^2. a = 120x2/(6.32)^2 = 6m/s^2.
Answer : The approximate acceleration of the car is, [tex]6.008m/s^2[/tex]
Solution : Given,
Initial velocity of the car = 0 m/s
Distance covered = 120 meter
Time taken = 6.32 second
Using second law of motion,
[tex]s=ut+\frac{1}{2}at^2[/tex]
where,
s = distance covered
u = initial velocity
t = time taken
a = acceleration
Now put all the given values in the above equation, we get
[tex](120m)=(0m/s)\times (6.32s)+\frac{1}{2}\times a\times (6.32s)^2[/tex]
[tex]a=6.008m/s^2[/tex]
Therefore, the approximate acceleration of the car is, [tex]6.008m/s^2[/tex]