Answer:
acceleration: 5.48 m/s²
time taken: 9.3 seconds
Given:
[tex]\boxed{\sf acceleration = \dfrac{Force}{mass}}[/tex] [tex]\boxed{\sf time \ taken = \dfrac{final \ velocity - initial \ velocity}{acceleration} }[/tex]
using the formula's:
answer 1:
[tex]\sf acceleration = \dfrac{F}{m}[/tex]
[tex]\sf acceleration = \dfrac{6572 }{1200 }[/tex]
[tex]\sf acceleration = 5.4766... \ m/s^2[/tex]
[tex]\sf acceleration = 5.48 \ m/s^2[/tex] // rounded to nearest hundredth //
answer 2:
[tex]\sf time \ taken = \dfrac{\Delta v}{a}[/tex]
[tex]\sf time \ taken = \dfrac{66 - 15}{5.48}[/tex]
[tex]\sf time \ taken = 9.3122 \ seconds[/tex]
[tex]\sf time \ taken = 9.3 \ seconds[/tex] // rounded it to nearest tenth //
Answer:
[tex]\large\boxed{\sf Time =9.30\ s }[/tex]
[tex]\large\boxed{\sf Accl^n =5.47\ m/s^2 }[/tex]
Explanation:
Here it is given that a car of mass 1200kg has an initial velocity of 15m/s achieves a velocity of 66m/s . A force of 6572N was applied in order to increase the velocity and we need to find the time taken in doing so .
From Newton's Second Law of Motion ,
[tex]\sf\qquad\longrightarrow Force = \triangle p\\\\ [/tex]
As momentum (p) = mass × velocity ,
[tex]\\\sf\qquad\longrightarrow Force = \dfrac{m(v-u)}{t} \\\\ [/tex]
Here ,
• On substituting the respective values ,
[tex]\sf\qquad\longrightarrow 6572N = \dfrac{1200kg ( 66- 15 )m/s }{t}\\\\ [/tex]
[tex]\sf\qquad\longrightarrow 6572N = 1200kg \times \dfrac{51}{t} \\\\ [/tex]
[tex]\sf\qquad\longrightarrow t = \dfrac{1200×51}{6572N } \\\\ [/tex]
[tex]\sf\qquad\longrightarrow t =\dfrac{61200}{6572}s \\\\ [/tex]
[tex]\sf\qquad\longrightarrow \pink{\frak{ Time = 9.30\ s }} [/tex]
• For finding acceleration ,
[tex]\sf\qquad\longrightarrow Force = mass * acceleration \\\\ [/tex]
[tex]\sf\qquad\longrightarrow accl^n =\dfrac{Force}{mass} \\\\ [/tex]
[tex]\sf\qquad\longrightarrow accl^n =\dfrac{6572N }{1200kg} \\\\ [/tex]
[tex]\sf\qquad\longrightarrow \pink{\frak{ acceleration= 5.47\ m/s^2}} \\\\ [/tex]
[tex]\rule{200}4[/tex]
