Suppose that a car starts from rest at position -3.31m and accelerates with a constant acceleration of 4.15m/s^2. At what time t is the velocity of the car 19.2m/s?
v=u+at
19.2=0+4.15t
t=19/4.15
t=4.8
check the answer
What is the "u" variable standing for?
Your answer's wrong, btw, sorry :(
u is the initial velocity that is 0 as it as rest ..
whats the answer ?
Thread necro... but I wanted to make sure that anyone seeing this can correctly divide 19.2/4.15 = 4.626506024
To find the time at which the velocity of the car is 19.2 m/s, we can use the kinematic equation:
v = u + at
Where:
- v is the final velocity (19.2 m/s),
- u is the initial velocity (0 m/s since the car starts from rest),
- a is the acceleration (4.15 m/s^2), and
- t is the time we want to find.
Since the car is starting from rest, the initial velocity u is 0 m/s.
Plugging in the given values, the equation becomes:
19.2 m/s = 0 m/s + (4.15 m/s^2)t
Simplifying the equation, we have:
19.2 m/s = 4.15 m/s^2 * t
Now, we can solve for t by rearranging the equation:
t = 19.2 m/s / 4.15 m/s^2
Calculating the value, we get:
t ≈ 4.63 seconds
Therefore, the time at which the velocity of the car is 19.2 m/s is approximately 4.63 seconds.