Sports Car Acceleration Question

AI Thread Summary
A sports car traveling at a constant speed of 22 m/s covers 110 meters in 5 seconds before braking to a stop in 4 seconds. To find the acceleration, the kinematic equation v = u + at is used, where the initial velocity (u) is 22 m/s and the final velocity (v) is 0 m/s. The calculation shows that the acceleration is -5.5 m/s², which converts to approximately -0.56 g's. The negative sign indicates deceleration as the car comes to a stop. The solution highlights the importance of understanding initial and final velocities in kinematic equations.
Adrianna
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Homework Statement


A sports car moving at constant speed travels 110m in 5.0s. If it then brakes and comes to a stop in 4.0s. what is its acceleration in m/s^2? Express the answer in terms of "g's." where 1.00 g = 9.80 m/s^2.


Homework Equations


I thought about using the equation a=v/t

The Attempt at a Solution


So I tried it sort of I am just sort of lost I am not really sure how to start...
 
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Adrianna said:

Homework Equations


I thought about using the equation a=v/t

You need to use the kinematic equations

v=u+at

v2=u2+2as

s=ut+1/2at2


Which one do you think you will have to use?
 
What does the u stand for, that's not in my book?
 
The car initially had a velocity of 110m/5s (22m/s) and came to a stop (0m/s) in 4.0s. The acceleration is the change in velocity over a time interval.
 
So would we use the v=u+at since I have the 4.0s and the 22m/s? but what would go in for the u?
 
u is initial velocity and v is final velocity.
 
Okay I figured it out thanks so much the answer was -5.5 m/s^2 and then in g's it was -0.56 g's
 
take u (initial speed) = 110 / 5 = 22m/s
v = 0
t = 4

v = u + at
0 = 22 + 4a
4a = -22
a = -5.5m/s^2
a = (-5.5/9.8)g = -0.56g
 
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