Question

Find the acceleration and distance covered by the car if the car accelerator uniformly from 5m/s² to 10m/s² in 5 second

Answer 1

Answer:

Initial velocity of the car (u)= 5 m/s

Final velocity of the car (v) = 10 m/s

Time taken for change in velocity (t)= 5 s

Therefore acceleration (a)= [10 m/s - 5 m/s]/ 5 s = 5 m/s /5s = 1 m/s²

Distance covered during these 5 seconds, is found using the relation

s = u t + ½ a t² = 5 m/s× 5 s + ½ 1 m/s² × 5² s²=25 m+ 12.5 m = 37.5 m

So car covers 37.5 m during the period its velocity changes from 5 m/s to 10 m/s.

Answer 2

$\huge\purple{ \underline{ \boxed{\mathbb{\red{QUESTION : }}}}}$

Find the acceleration and distance covered by the car if the car accelerates uniformly from 5m/s to 10m/s in 5 second .

$\huge\purple{ \underline{ \boxed{\mathbb{\red{ANSWER : }}}}}$

• Acceleration of the car = 1 m/$\tt {s}^{2}$

• Distance Covered by car = 37.5 m

$\huge\purple{ \underline{ \boxed{\mathbb{\red{EXPLANATION : }}}}}$

$\green{ \large \underline{ \mathbb{\underline{GIVEN : }}}}$

• Initial velocity ( u ) = 5 m/s

• Final velocity ( v ) = 10 m/s

• Time taken ( t ) = 5 s

$\green{ \large \underline{ \mathbb{\underline{TO \: FIND : }}}}$

• Acceleration of the car .

• Distance covered by the car.

$\green{ \large \underline{ \mathbb{\underline{ EQUATIONS \: OF \: MOTION \: USED : }}}}$

$\large{\purple{ \boxed{\begin{array}{l} \sf v = u + at \\ \\ \sf {v}^{2} - {u}^{2} = 2as \: \: \end{array}}}}$

$\green{ \large \underline{ \mathbb{\underline{SOLUTION: }}}}$

$\large{\purple{ \underline{\underline{\textsf{Acceleration of the car}}}}}$

$\displaystyle \sf \longrightarrow v = u + at \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow 10 = 5 + a(5) \\\ \\ \displaystyle \sf \longrightarrow 10 = 5 + 5a \: \: \: \\ \\ \displaystyle \sf \longrightarrow 5 + 5a = 10 \: \: \: \\ \\ \displaystyle \sf \longrightarrow 5a = 10 - 5 \: \: \\ \\ \displaystyle \sf \longrightarrow 5a = 5\: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow a = \frac{5}{5} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow a = 1 \: m {s}^{ - 2} \: \: \: \: \:$

$\large{\purple{ \underline{\underline{\textsf{Distance Covered by car}}}}}$

$\: \: \displaystyle \sf \longrightarrow {v}^{2} - {u}^{2} = 2as \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow {(10)}^{2} - {(5)}^{2} = 2(1)s \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow 100 - 25 = 2s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\\displaystyle \sf \longrightarrow 75 = 2s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow 2s = 75 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow s = \frac{75}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow s = 37.5 \: m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple{\boxed{ \begin{array}{l} \textsf{Acceleration of the car = 1 m { \sf s}^{ - 2} } \\ \\ \textsf{Distance Covered by car = 37.5 m} \end{array}}}$

$\green{ \large \underline{ \mathbb{\underline{KNOW\:MORE: }}}}$

• Acceleration

Acceleration is the rate at which velocity changes with time .

• Initial Velocity

Initial velocity is the velocity of the object before the effect of acceleration .

• Final Velocity

Final velocity is the velocity of the object after the effect of acceleration .

• Distance

Distance is the length of actual path covered by a moving object in a given time interval .

$\Large{\purple{\boxed{\begin{array}{l} \textsf{Equations of motion : } \\ \\ \textsf{v = u + at} \\ \\ \displaystyle\textsf{s = ut + \sf\frac{1}{2}a {t}^{2} } \\ \\ \sf {v}^{2} - {u}^{2} = 2as \end{array}}}}$