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

We have to find acceleration first.

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

$\boxed{\sf Acceleration=\dfrac{v-u}{t}}$

$\\ \sf\longmapsto Acceleration=\dfrac{22-44}{11}$

$\\ \sf\longmapsto Acceleration=\dfrac{-22}{11}$

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

• Now
• Distance=s

Using second equation of kinematics

$\boxed{\sf s=ut+\dfrac{1}{2}at^2}$

$\\ \sf\longmapsto s=44(11)+\dfrac{1}{2}(-2)(11)^2$

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

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

$\\ \sf\longmapsto s=363m$

Answer 2

Explanation:

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