A particle has a velocity of 18m/s at a certain time and 2.4s later is velocity is 30m/s in the opposite direction. What are the magnitude and direction of the particle's acceleration during the 2.4s interval?
I udes a=change in velocity /change in time and I got -20m/s^2 is this correct?
If we use
(v_f - v_o)/t = a where v_f=-30m/s, v_o=18m/s and t=2.4s then
a = (-30-18)/2.4 m/s^2
So yes, it looks correct to me.
-20m/s is the acceleration
Yes, you are correct. To calculate the acceleration during the 2.4s interval, you can use the formula:
a = (v_f - v_i) / t
where a is the acceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time interval.
In this case, the final velocity is -30m/s (opposite direction), the initial velocity is 18m/s, and the time interval is 2.4s. Plugging these values into the formula, we get:
a = (-30 - 18) / 2.4 m/s^2
a = -48 / 2.4 m/s^2
a = -20 m/s^2
So, the magnitude of the particle's acceleration during the 2.4s interval is 20 m/s^2, and since the result is negative, it tells us that the acceleration is in the opposite direction to the initial velocity.