the equation for the motion of an object with constant acceleration is d=vt+0.5at^2where d is the distance travelled in ft, v is the starting velocity in ft/s a is the acceleration in ft/s^2 and t is time in seconds
a race car begins with a velocity of 200ft/s and accelerates at 100ft/s
what is the expression for the distance travelled after one sec
well, how about using the numbers given, in the formula given?
d=200t+50t^2
To find the expression for the distance traveled after one second, we'll substitute the given values into the equation for motion of an object with constant acceleration:
d = vt + 0.5at^2
Given:
v = 200 ft/s (starting velocity)
a = 100 ft/s^2 (acceleration)
t = 1 sec (time)
Let's substitute the values into the equation:
d = (200 ft/s)(1 sec) + 0.5(100 ft/s^2)(1 sec)^2
Simplifying,
d = 200 ft + 0.5(100 ft/s^2)(1 sec)^2
d = 200 ft + 0.5(100 ft/s^2)(1 sec)
d = 200 ft + 0.5(100 ft/s^2)
d = 200 ft + 0.5(100 ft)
d = 200 ft + 50 ft
d = 250 ft
Therefore, the expression for the distance traveled after one second is d = 250 ft.
To find the expression for the distance traveled after one second, we can use the equation for the motion of an object with constant acceleration:
d = vt + 0.5at^2
Given:
- Starting velocity (v) = 200 ft/s
- Acceleration (a) = 100 ft/s^2
- Time (t) = 1 second
We can substitute these values into the equation to find the expression for the distance traveled after one second:
d = (200 ft/s)(1 s) + 0.5(100 ft/s^2)(1 s)^2
First, let's calculate the term vt:
vt = (200 ft/s)(1 s)
= 200 ft
Next, let's calculate the term 0.5at^2:
0.5at^2 = 0.5(100 ft/s^2)(1 s)^2
= 0.5(100 ft/s^2)(1 s^2)
= 0.5(100 ft)
= 50 ft
Now, let's substitute these values back into the equation for d:
d = vt + 0.5at^2
= 200 ft + 50 ft
= 250 ft
Therefore, the expression for the distance traveled after one second is d = 250 ft.