Question

a car moves uniformly from 5m/s to 10m/s in 5 sec. calculate the acceleration and distance covered.​

Answer 1

a= v-u/ t

here,

v=10m/s

u=5 m/s

t= 5 s

hence

a=10-5/5 m/s^2

a=5/5

a=1 m/s^2

applying 3rd equation to find s

v^2-u^2=2as

100-25=2(1×s)

75=2s

s=37.5 m

Answer 2

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

A car accelerates uniformly from 5m/s to 10m/s in 5s . Calculate the acceleration and the distance covered be the car in that time .

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

• 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 .

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