The speed of a car traveling on a straight road increases from 63 m/s to 75
m/s in 4.2 seconds. What was the car's acceleration? m/s2 = meters per
second squared

Answers

Answer 1
Answer :

Initial velocity = 63m/s

Final velocity = 75m/s

Time taken = 4.2

We have to find acceleration of car[tex].[/tex]

_________________________________

◈ Acceleration is defined as the rate of change of speed.

a = (v - u) / t

It is a vector quantity having both magnitude as well as direction.

SI unit : m/s²

⇒ a = (v - u) / t

⇒ a = (75 - 63) / 4.2

⇒ a = 12/4.2

⇒ a = 2.85 m/s²
Answer 2
Question:

The speed of a car traveling on a straight road increases from 63 m/s to 75 m/s in 4.2 seconds. What was the car's acceleration?

Answer:

[tex] \rm\red {Given:} [/tex]

u ( initial velocity ) = 63m/s

v ( final velocity ) = 75m/s

t ( time taken ) = 4.2s

[tex] \rm\red {To \: find:} [/tex]

a ( acceleration ) =?

[tex] \rm\red {Calculation:} [/tex]

We know that,

[tex] {\huge {\boxed {\tt { a = \dfrac{v-u} {t}}}}} [/tex]

Substituting values-

[tex] \sf { \qquad \leadsto a = \dfrac{75-63}{4.2}} [/tex]

[tex] \sf { \qquad \leadsto a = \dfrac{12}{4.2}} [/tex]

[tex] \sf { \qquad \leadsto a = \dfrac{\cancel {12} {} ^{\: \: 6} \times 10}{\cancel {42} {} ^{\: \: 21} }} [/tex]

[tex] \sf { \qquad \leadsto a = \dfrac{\cancel6 {} ^{\: \: \: 2}\times 10 }{\cancel {21} {} ^{\: \: 7}}} [/tex]

[tex] \sf { \qquad \leadsto a = \dfrac{2 \times 10}{7}} [/tex]

[tex] \sf { \qquad \leadsto a = \dfrac{20}{7} = 2.85} [/tex]

[tex] \therefore [/tex] Acceleration of the car was [tex] \sf { 2.8m/{s} ^{2}} [/tex]

_________________________________

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Answers

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