Answer:
a) a= -1.5 m/s²
b) t = 4.0 s
Explanation:
a)
- Since we are told that the deceleration is uniform, we can use the following kinematic equation:
[tex]v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta X (1)[/tex]
- Since the truck finally stops, this means that vf = 0.
- Replacing the values of v₀, and Δx in (1), we can solve for a, as follows:
[tex]a =\frac{-(v_{o}^{2})}{2*\Delta X} = \frac{-(6.0m/s)^{2} }{2* 12.0m} = -1.5 m/s2 (2)[/tex]
b)
- Since we know the value of the initial and final velocity, and the value of the acceleration also, we can apply the definition of acceleration, solving for time t, as follows:
[tex]\Delta t = \frac{-v_{o}}{a} = \frac{-6.0m/s}{-1.5m/s2} = 4.0 s (2)[/tex]
