Answer:
57.19461 m/s²
-9.20833 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{663}{3.6}-0}{3.22}\\\Rightarrow a=57.19461\ m/s^2[/tex]
The acceleration during the race is 57.19461 m/s²
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-\dfrac{663}{3.6}}{20}\\\Rightarrow a=-9.20833\ m/s^2[/tex]
The acceleration while stopping is -9.20833 m/s²
The assumptions made here are:
The initial velocity of the dragster at the start of the race is zero
The acceleration is constant in both cases of acceleration.
The final velocity when it stops is zero
Motion is in a straight line path