Question

Velocity v= 2t²+7t. find acceleration and displacement at time=5 seconds.

please answer Asap​

Answer 1

AnsWer :-

• v = 2t² + 7t

Acceleration :

$\longrightarrow \sf a = \dfrac{dv}{dt}$

$\longrightarrow \sf a = \dfrac{d}{dt}( {2t}^{2} + 7t)$

$\longrightarrow \sf a =4t + 7$

$\longrightarrow \sf a_{(t = 5 \: s)} =4t + 7$

$\longrightarrow \sf a_{(t = 5 \: s)} =4(5)+ 7$

$\longrightarrow \sf a_{(t = 5 \: s)} =20 + 7$

$\longrightarrow \underline{ \boxed{ \bold{ a_{(t = 5 \: s)} =2 7 \: {ms}^{ - 2} } }}$

Displacement :

$\longrightarrow \sf v = \dfrac{dx}{dt}$

$\longrightarrow \sf dx = v.dt$

Integrating both the sides we have:

[tex]\longrightarrow \:\displaystyle \sf \int\limits_{0}^{x} dx = \int\limits_{0}^{5}v.dt \\ [/tex]

[tex]\longrightarrow \:\displaystyle \sf \int\limits_{0}^{x} dx = \int\limits_{0}^{5}( {2t}^{2} + 7t).dt \\ [/tex]

[tex]\longrightarrow \:\displaystyle \sf x = \bigg( \left. \dfrac{ {2t}^{3} }{3} + \frac{7 {t}^{2} }{2} \bigg) \right| _{0}^{5} \\ [/tex]

[tex]\longrightarrow \:\displaystyle \sf x = \Bigg[ \dfrac{ {2(5)}^{3} }{3} - \dfrac{ {2(0)}^{3} }{3}\Bigg] +\Bigg[ \dfrac{7 {(5)}^{2} }{2} - \dfrac{7 {(0)}^{2} }{2}\Bigg][/tex]

[tex]\longrightarrow \:\displaystyle \sf x = \Bigg[ \dfrac{2 \times 125}{3} - 0\Bigg] +\Bigg[ \dfrac{7 \times 25}{2} - 0\Bigg][/tex]

[tex]\longrightarrow \:\displaystyle \sf x = \dfrac{250}{3} + \dfrac{175}{2} [/tex]

[tex]\longrightarrow \:\displaystyle \sf x = \dfrac{2 \times 250 + 3 \times 175}{6} \\ [/tex]

[tex]\longrightarrow \:\displaystyle \sf x = \dfrac{500 + 525}{6} \\ [/tex]

[tex]\longrightarrow \:\displaystyle \sf x = \dfrac{1025}{6} \\ [/tex]

[tex]\longrightarrow \: \underline{ \boxed{\displaystyle \bold{ x = 170.84 \: m }}}\\ [/tex]