Given:
The distance covered by truck: d = 40.0 m
The time taken to cover the distance is: t = 9.00 s
The final velocity of the truck is: v2 = 2.20 m/s
To find:
a) the speed of the truck.
b) the acceleration
Explanation:
a)
The speed of the truck before it slows down can be calculated as:
[tex]d=\frac{1}{2}(v_2+v_1)t[/tex]
Substituting the values in the above equation, we get
[tex]\begin{gathered} 40=\frac{1}{2}(2.20+v_1)\times9 \\ \\ \frac{40\times2}{9}-2.20=v_1 \\ \\ v_1=6.69\text{ m/s} \end{gathered}[/tex]
b)
The truck is initially moving at a speed of 6.69 m/s. It then slows down to the final velocity of 2.20 m/s. The acceleration of the truck can be determined as:
[tex]d=v_1t+\frac{1}{2}at^2[/tex]
Substituting the values in the above equation, we get:
[tex]\begin{gathered} 40=6.69\times9+\frac{1}{2}\times a\times9^2 \\ \\ 40=60.21+40.5a \\ \\ a=\frac{40-60.21}{40.5} \\ \\ a=-0.499 \\ \\ a\approx-0.5\text{ m/s}^2 \end{gathered}[/tex]
Final answer:
a) The original speed of the truck is 6.69 m/s.
b) The acceleration of the truck is - 0.5 m/s^2.
