what is the acceleration of a bowling ball that starts at rest and moves 300m down the gutter in 22.4 sec

1 answer:

7 0

<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>

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At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.