A sports car moving at constant speed travels 90m in 4.9s. If it then brakes and comes to a stop in 4.0s. Express the answer in terms of g's, where g= 9.8m/s^s
I tried this question and got 0.47g, but it was incorrect.
2007-02-14 13:42:47 · 3 answers · asked by Vanita L 1 in Science & Mathematics ➔ Physics
Your number is correct, the magnitude of the acceleration is 0.47g's. Maybe they want you to say the acceleration is -0.47 m/s/s, since the final velocity is 0 and the initial velocity is 90/4.9 m/s and a = (Vf-Vo)/t
2007-02-14 13:52:50 · answer #1 · answered by Dennis H 4 · 0⤊ 0⤋
Acceleration is the change in velocity. Velocity is the vector of speed and direction. This is why you are pulled to the outside of a turn. The speed is constant, but direction changes.
v=90m/4.9s
v=18.37m/s
a=v1-v2/t
a=90-0/4s
a=22.5m/s^2
22.5m/s^2 / 9.8m/s^2
2.3 g
2007-02-14 21:51:51 · answer #2 · answered by Matthew P 4 · 0⤊ 0⤋
After the car stops, it is imposing a 0G force. Let me be a little more clear. A car, no matter what it's constant speed is does not accelerate during braking.
2007-02-14 21:46:50 · answer #3 · answered by ttpawpaw 7 · 0⤊ 0⤋