Question

a car is moving a straight line with initial velocity 5m/s.after time 10 sec its velocity becomes 20m/s.calculate the acceleration. then apply brakes it comes to rest. after time 20s. calculate the acceleration i this case also​

Answer 1

$\LARGE\underline{\sf{\red{Required\:answer:-}}}$

CASE - I :

• Initial velocity of the car (u) = 5 m/s
• Time (t) = 10 sec
• Final velocity of the car (v) = 20 m/s

Using First equation of motion ,

$\\ \star \:\large\boxed{\purple{\sf{v = u + at}}}$

By substituting the values we have ,

$\\ : \implies \sf \: 20 \: m {s}^{ - 1} = 5 \: m {s}^{ - 1} + a(10 \: s)$

$\\ : \implies \sf \: 20 \: m {s}^{ - 1} - 5 \: m {s}^{ - 1} = a(10 \: s)$

$\\ :\implies \sf \: 15 \: m {s}^{ - 1} = a(10 \: s)$

$\\ : \implies \sf \dfrac{15 \:m{s}^{ - 1}}{10 \: s} = a$

$\\ : \implies \boxed{\pink{\mathfrak{a =1.5 \: m {s}^{ - 2} }}} \: \bigstar$

Hence , The accleeration of the car is 1.5 m/s²

━━━━━━━━━━━━━━━━━━━━━━━

CASE - II :

• Initial velocity of the car (u) = 5 m/s
• Time (t) = 20 sec
• Final velocity of the car (v) = 0 m/s [ since it came to rest]
• Acceleration (a) = ?

using First equation of motion ,

$\\ : \implies \sf \: 0 \: m {s}^{ - 1} = 5 \: m {s}^{ - 1} + a(20 \: s)$

$\\ : \implies \sf \: - 5 \: m {s}^{ - 1} = a(20 \: s)$

$\\ : \implies \sf \: a = \frac{ - 5 \: m {s}^{ - 1} }{20 \: s}$

$\\ : \implies{\boxed{\mathfrak{\pink{a = - 0.25 \: m {s}^{ - 2} }}}} \: \bigstar$

Hence , The acceleration of the car is - 0.25 m/s² . [ Negative sign indicates deceleration]

Answer 2

Answer:

\LARGE\underline{\sf{\red{Required\:answer:-}}}

Requiredanswer:−

CASE - I :

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

Time (t) = 10 sec

Final velocity of the car (v) = 20 m/s

Using First equation of motion ,

\begin{gathered}\\ \star \:\large\boxed{\purple{\sf{v = u + at}}} \\\end{gathered}

⋆

v=u+at

By substituting the values we have ,

\begin{gathered}\\ : \implies \sf \: 20 \: m {s}^{ - 1} = 5 \: m {s}^{ - 1} + a(10 \: s) \\ \\\end{gathered}

:⟹20ms

−1

=5ms

−1

+a(10s)

\begin{gathered}\\ : \implies \sf \: 20 \: m {s}^{ - 1} - 5 \: m {s}^{ - 1} = a(10 \: s) \\ \\\end{gathered}

:⟹20ms

−1

−5ms

−1

=a(10s)

\begin{gathered}\\ :\implies \sf \: 15 \: m {s}^{ - 1} = a(10 \: s) \\ \\\end{gathered}

:⟹15ms

−1

=a(10s)

\begin{gathered}\\ : \implies \sf \dfrac{15 \:m{s}^{ - 1}}{10 \: s} = a \\ \\\end{gathered}

:⟹

10s

15ms

−1

=a

\begin{gathered}\\ : \implies \boxed{\pink{\mathfrak{a =1.5 \: m {s}^{ - 2} }}} \: \bigstar \\ \\\end{gathered}

:⟹

a=1.5ms

−2

★

Hence , The accleeration of the car is 1.5 m/s²

━━━━━━━━━━━━━━━━━━━━━━━

CASE - II :

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

Time (t) = 20 sec

Final velocity of the car (v) = 0 m/s [ since it came to rest]

Acceleration (a) = ?

using First equation of motion ,

\begin{gathered}\\ : \implies \sf \: 0 \: m {s}^{ - 1} = 5 \: m {s}^{ - 1} + a(20 \: s) \\ \\\end{gathered}

:⟹0ms

−1

=5ms

−1

+a(20s)

\begin{gathered}\\ : \implies \sf \: - 5 \: m {s}^{ - 1} = a(20 \: s) \\ \\\end{gathered}

:⟹−5ms

−1

=a(20s)

\begin{gathered}\\ : \implies \sf \: a = \frac{ - 5 \: m {s}^{ - 1} }{20 \: s} \\ \\\end{gathered}

:⟹a=

20s

−5ms

−1

\begin{gathered}\\ : \implies{\boxed{\mathfrak{\pink{a = - 0.25 \: m {s}^{ - 2} }}}} \: \bigstar \\ \\\end{gathered}

:⟹

a=−0.25ms

−2

★

Hence , The acceleration of the car is - 0.25 m/s² . [ Negative sign indicates deceleration]