Question

The truck travels in a circular path having a radius of 50 m at a speed of v = 4 m>s. For a short distance from s = 0, its speed is increased by v # = (0.05s) m>s 2, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s = 10 m.

Answer 1

Answer:

Magnitude of velocity when s is 10 m= 4.58m/s

Magnitude of acceleration when s is 10 m = 0.653 m/s²

Explanation:

Radius of circular path = 50m

Speed = 4m/s

increase in acceleration = (0.05s)m/s²

speed and acceleration when s is 10 m = ?

Magnitude of Velocity:

From third equation of motion

$v_{f}^{2}-v_{i}^{2}=2aS\\\\v_{f}=\sqrt{2aS+v_{i}^{2}} \\\\v_{f}=\sqrt{2(0.05)(10)+(4)^{2}} \\\\v_{f}=4.58\,m/s$

Magnitude of Acceleration:

$a=\sqrt{a_{t}^{2}+a_{n}^{2}}--(1)$

The tangential component of acceleration is

$a_{t}=(0.05)(10)\\a_{t}=0.5\,m/s^{2}$

The normal component of acceleration is

$a_{n}=\frac{v^{2}}{R}\\\\a_{n}=\frac{16}{50}\\\\a_{n}=0.42\,m/s^{2}$

Substituting tangential and normal component in (1)

$a=\sqrt{(0.5)^{2} +(0.42)^{2} }$

a=0.653 m/s²