Question: Kinematics Question ( Physics )?

At a car sports rally, a car starting from
rest accelerates uniformly at a rate of 9.0
m/s over a straight-line distance of 100
m. The time to beat is 4.5 s. Does the
driver do it? If not what must the
acceleration be to do so?

thanks guys!!!

Relation Questions:

x=(v1+v2):2 .t => 100=(0+9):2 .t => t=22.2 IT IS MORE THAN 4.5 s!

Hi.
X=100, V0=9 , t=4.5 .
X=V*t+X0 and X0=0,
100=9*4.5 and this is not true and he should change his rate.
100=V*4.5 then V= 200/9 .
V=a*t+V0 ; 200/9=a*4.5+9 then a=2.93 m/s^2 .
:)
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distance(as a function of time) = starting position + starting velocity * t + ½* acceleration * t²

100 = 0 + 0t + ½*a*(4.5)²
100 = ½*a*20.25
a = 100/(10.125) = 9.87654321 m/sec²

so, the acceleration was not fast enough

check:

a * t = v
9.87654321 * 4.5 = 44.444 m/sec at the end of 4½ sec

average = 22.225 m/sec
distance = 4½ sec * 22.225 m/sec = slightly over 100 m

and

½*a*t² = ½ * 9.87654321 * (4.5)² = just over 100 m

x(f)=x(i)+v(i)t+1/2at^2

x(f)=unknown
x(i)=0
v(i)=o
t=4.5
a=9

plug into equation and you get 91 M. So nope.

s=100, V0=9 , t=4.5 .
s=V*t+X0 and X0=0,
100=9*4.5 and this is not true and he should change his rate.
100=V*4.5 then V= 200/9 .
V=a*t+V0 ; 200/9=a*4.5+9 then a=2.93 m/s^2 .

no he will not made it .