A truck gradually starts off from rest with a uniform acceleration of 2m\s It reaches a velocity of 16m\s calculate the distance lttravels lt is accelerating?
V^2 = Vo^2 + 2a*d.
16^2 = 0 + 4*d,
d = 64 m.
acceleration time = velocity / acceleration ... this is t
acceleration distance = 1/2 * a * t^2
Please healp me!!
To calculate the distance that the truck travels while accelerating, you can use the kinematic equation:
\[v^2 = u^2 + 2as\]
where:
v = final velocity (16 m/s)
u = initial velocity (0 m/s, as the truck starts from rest)
a = acceleration (2 m/s^2)
s = distance traveled
First, let's rearrange the equation to solve for s:
\[s = \frac{v^2 - u^2}{2a}\]
Plugging in the given values into the equation, we get:
\[s = \frac{(16^2) - (0^2)}{2 \times 2}\]
\[s = \frac{256}{4}\]
\[s = 64 \, \text{m}\]
Therefore, the distance traveled by the truck while accelerating is 64 meters.