Question

Y UD Eda

NF-22
11. A race car traveling at +44 m/s is uniformly accelerated to a velocity of +22 m/s over an
11-s interval. What is its displacement during this time?
a

Answer 1

Step-by-step explanation:

`we \: want \: to \: find \: displacement \\ we \: can \: use \: newtons \: laws \: of \: motion \\ s = ut + (1)/(2) a {t}^(2)....(1) \\here \: 44 = u \: (initial) \\ v = 22(final) \\ t = 11 \\ then \: \\ s = 44 * 11 + (1)/(2) ((22 - 44)/(11) ) {11}^(2) \\ s = 484 + (1)/(2) * ( - 22)/(11) * 11 * 11 \\ s = 484 - 121 \\ s = 363m \\ thank \: you`

Answer 2

We have to find acceleration first.

• Initial velocity=u=44m/s
• Final velocity=v=22m/s
• Time=11s=t

`\boxed{\sf Acceleration=(v-u)/(t)}`

`\\ \sf\longmapsto Acceleration=(22-44)/(11)`

`\\ \sf\longmapsto Acceleration=(-22)/(11)`

`\\ \sf\longmapsto Acceleration=-2m/s^2`

• Now
• Distance=s

Using second equation of kinematics

`\boxed{\sf s=ut+(1)/(2)at^2}`

`\\ \sf\longmapsto s=44(11)+(1)/(2)(-2)(11)^2`

`\\ \sf\longmapsto s=484+(-1)(121)`

`\\ \sf\longmapsto s=484-(121)`

`\\ \sf\longmapsto s=363m`