[填空题]
An object moves in a stldn vc63y b e:mgr2(,j1zrb*uraight line on a level road. The vari-y xqhas5 ma*08fo r +8a8ffbcation of the object's distance d with m8-saffa+c8 5 * qab8oxhfy r0time t is shown on the graph below
1. During the interval of $t=0.5\,s$ and $t=1.0\,s$, the object
2. Calculate the instantaneous speed of the object at $t=0.5\,s$ the answer is $v$ = $m/s$
3. Determine the direction of the change in momentum of the object during the motion.
4. On the axes below, sketch a possible graph of the variation of velocity v of the object with time t. There is no need to add values to the axes.
参考答案: 空格1: slows down
空格2: $18±2$
空格3: negative
本题详细解析:
1. The slope of the graph gives the speed of the object. The slope decreases between $t=0.5\,s$ and $t=1.0\,s$. Therefore the object slows down.
2. Any evidence for drawing tangent at $t = 0.5\,s$. $\frac{28 - 10}{1 - 0} \Rightarrow 18 \pm 2 \text{ m s}^{-1}$
Further Explanation
The gradient of the distance-time graph indicates the speed of the object. In one-dimensional motion, speed and velocity have the same magnitude. Since the gradient gets a smaller value with time and ends up with a zero value, velocity gets smaller and be zero at the end.The exact mathematical shape of the distance-time curve is not known, therefore it is not possible to determine the exact shape of the velocity-time graph. A linear graph has been included here, but curvature in the graph is permitted.
