A car traveling at 22.0 m/s accelerates for 2.37 seconds at a rate of 2.25 m/s2. How far will it travel in meters during this time?
(22.0)(2.37) +1/2(2.25)(2.37)= answer 58.5m right?
d = Vi t + (1/2) a t^2
(22.0)(2.37) +1/2(2.25)(2.37)^2 = 52.14 + 6.32 = 58.5
so I agree with your calculator but not your formula :)
t^2 !!!!!! t times t
I did calc t^2 didn't write it though
d=Vsub0t +1/2at^2
To find the distance traveled by a car during a given time interval, we can use the formula:
d = v₀t + 1/2at²
where:
d = distance traveled
v₀ = initial velocity
t = time
a = acceleration
Given values:
v₀ = 22.0 m/s (initial velocity)
t = 2.37 s (time)
a = 2.25 m/s² (acceleration)
Plugging in the values, we get:
d = (22.0 m/s)(2.37 s) + 1/2(2.25 m/s²)(2.37 s)