I just want some help checking my homework...
Question: A care travelling at a velocity of 42ms-1 suddenly loses engine power and slows at a constant rate. It comes to rest after 75 seconds.
a what is the acceleration during this time ?
b how far did the care travel before stopping ?
I used v== v + at ...and came up with - 0.56 m/s-2
and 1533 metres travelled ???
Help please. I have no idea if I am even remotely close to the right answer.
yes, 0 = 42 + 75 a
a = - .56
d = Vi t + .5 a t^2
d = 42(75) - .28(75)^2
= 3150 - 1575 = 1575
Thank you so much.... I got that wrong as didn't know whether to use the time... Got confused as the car was already moving .
You are welcome.
To find the acceleration, you can use the formula:
v = u + at,
where
v = final velocity (0 m/s, as the car comes to rest)
u = initial velocity (42 m/s)
t = time taken (75 s)
a = acceleration (unknown)
Rearranging the formula:
0 = 42 + a * 75.
Solving for a:
a = (0 - 42) / 75.
a = -42 / 75.
a = -0.56 m/s².
So, your answer for the acceleration (-0.56 m/s²) is correct!
Now, let's find the distance traveled before the car stopped. You can use the formula:
s = ut + (1/2)at²,
where
s = distance traveled (unknown)
u = initial velocity (42 m/s)
t = time taken (75 s)
a = acceleration (-0.56 m/s²).
Plugging in the values and solving:
s = (42 * 75) + (0.5 * -0.56 * 75²).
s = 3150 - 1575.
s = 1575 meters.
So, the correct answer for the distance traveled is 1575 meters. Therefore, it seems like you've got it right!