Question

A car of mass 1000kg travelling at 72km/h uniformily brought to rest over a distance of 40 m. find the average acceleration?​

Answer 1

Answer:

a = -5 m/s^2

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Explanation:

72km/h = 20m/s

then

use the formula

v_final^2 = v_initial^2+2.a.distance

0 = 400 m^2/s^2 +2.a.40m

80m . a = -400 m^2/s^2

a = -5 m/s^2

Answer 2

$\large{\underline{\underline{\mathfrak{Answer :}}}}$

• Acceleration is -5 m/s²

$\rule{200}{0.5}$

$\underline{\underline{\mathfrak{Step-By-Step-Explanation :}}}$

Given :

• Mass of car (m) = 1000 kg
• Initial velocity (u) = 72 km/h = 20
• Final velocity (v) = 0 km/h
• Distance (s) = 40 m

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To Find :

• Acceleration

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Solution :

Use 3rd equation of motion :

$\large{\boxed{\sf{v^2 \: - \: u^2 \: = \: 2as}}} \\ \\ \implies {\sf{0 \: - \: 20^2 \: = \: 2 \: \times \: a \: \times \: 40}} \\ \\ \implies {\sf{-400 \: = \: 80a}} \\ \\ \implies {\sf{a \: = \: \dfrac{-400}{80}}} \\ \\ \implies {\sf{a \: = \: -5}} \\ \\ \underline{\sf{\therefore \: Acceleration \: is \: -5 \: ms^{-2}}}$