If a racecar accelorates from 0-330 mph in 4 seconds how quickly does it reach 60 mph?

Answer 1

In both cases initial speed = 0 mph.
acceleration = 330 mph / 4 s = 60 mph / t
t x 330 mph = 4 s x 60 mph
t = 0.73 s
Th

Answer 2

No idea; cars don't accelerate at a constant velocity, so you can't get an answer more accurate than less that 4 seconds.

I would suggest that if the vehicle can attain 330 mph then it would reach 60 mph very quickly.

Answer 3

racecar accelorates from 0-330 mph in 4 seconds
So, 330mph = 4 sec (or) 4sec / 330mph = 1 ---> (1)
For 60mph,
60mph = x sec ---> (2)
Multiply eqn(2) by 5.5(This is done for equating the two equations)
So,eqn (2) become,
330 mph = 5.5x sec (or) 5.5x sec/ 330 mph=1 --->(3)
Now equating (1) and (3),
4 sec/330 mph = 5.5x sec/ 330 mph
(or)

5.5x sec = 4*330/330
(or)

x sec = 4/5.5

x sec = 0.72727273 sec

Therefore, it takes 0.73(approximately) to reach 60mph....

Answer 4

Just a thought... If the car accelerates at the indicated rate, the average, not instantaneous, acceleration would be 3.76 G's. If the acceleration is non-constant, then at some time the instantaneous acceleration must be greater than the average. If gLOC occurs at 6-8 G's, let's hope that your driver is wearing a flight suit as he's going to have tunnel vision at the least and maybe even a brownout.

Answer 5

It is simply 60/(330/4)=0.7272

Answer 6

Roughly 3/4 sec.

Answer 7

If accelleration is constant (which it isn't in real life) v = u + at, or
a = (v - u)/t = 330/4 = 82.5
using same eq again, t = (v - u)/a = 60 / 82.55

Answer 8

converting 330mph in metre/s as the time is in seconds
330mph=134m/s(approvimately)
60mph=27m/s(approximately)
Now,

v=u+at
134=0+a*4
a=134/4
a=33.5m/s^2

Time taken to reach 60mps or 27m/s
v=u+at
27=0+33.5*t
t=27/33.5
t=0.805..........seconds

Answer 9

.72secs