A car traveling at 80km/hr decelerates at 1.5m/s^2. Calculate
a) the time it takes to stop.
b) The distance travelled in the 3rd second.
a. Vf=Vi-at solve for t.
b. find distance at t=3, then distance at t=4 and subtract
d=vi+at where t=3, then do it for t=4
convert 80km/hr to m/s first.
b)
d=vi+at
d=22.22+1.5+3=26.72
d=22.22+1.5+4=27.72
d=27.72-26.72
d=1
is it correct for
To calculate the time it takes for the car to stop, we need to find the deceleration time.
a) The deceleration is given as 1.5 m/s^2. We need to convert the car's speed from kilometers per hour (km/hr) to meters per second (m/s), as the deceleration is given in meters per second squared (m/s^2).
To convert km/hr to m/s, we use the conversion factor: 1 km/hr = 1000 m/3600 s = 5/18 m/s.
So, the car's speed in m/s is 80 km/hr × (5/18) m/s = 200/9 m/s.
Now, we can use the formula:
v = u + at
Where:
v = final velocity (0 m/s, as the car comes to a stop)
u = initial velocity (200/9 m/s)
a = acceleration/deceleration (-1.5 m/s^2)
t = time taken to stop (unknown)
Plugging in the values, we have:
0 = 200/9 + (-1.5)t
Solving for t, we get:
-1.5t = -200/9
t = (-200/9) / (-1.5)
t ≈ 14.81 seconds (rounded to 2 decimal places)
Therefore, it takes approximately 14.81 seconds for the car to stop.
b) To find the distance traveled in the third second, we can use the formula:
s = ut + (1/2)at^2
Where:
s = distance traveled (unknown)
u = initial velocity (200/9 m/s)
a = acceleration/deceleration (-1.5 m/s^2)
t = time (3 seconds)
Plugging in the values, we have:
s = (200/9) × 3 + (1/2) × (-1.5) × (3^2)
s = 600/9 - (9/2)
s = (200 - 81)/3
s = 119/3 ≈ 39.67 meters (rounded to 2 decimal places)
Therefore, the distance traveled in the third second is approximately 39.67 meters.