Question

A race car is traveling at +50 m/s when is slows down at -10 m/s2 for 4seconds. What is his new velocity?

Answer 1

Given :

• Initial velocity of Car, u = 50 m/s
• Acceleration of Car, a = -10 m/s²
• Time for which care decelerates, t = 4 s

To find :

• Final velocity of Car, v =?

Formula required :

• First equation of motion

     v = u + a t

[ Where v is final velocity, u is initial velocity, a is acceleration and t is time taken ]

Calculation :

Using first equation of motion

→ v = u + a t

→ v = ( 50 ) + ( - 10 ) ( 4 )

→ v = 50 - 40

→ v = 10 m/s

Therefore,

• Final velocity of Car is 10 m/s.

Three equation of motion :

• First equation of motion

     v = u + a t

• Second equation of motion

     s = u t + 1/2 a t²

• Third equation of motion

     2 a s = v² - u²

[ Where v is final velocity, u is initial velocity, a is acceleration, t is time taken and s is distance covered ]

Answer 2

Answer:

✡ Given ✡

$\leadsto$ A race car is traveling at 50 m/s when is slows down at -10 m/s for 4 sec.

✡ To Find ✡

$\leadsto$ What is the final velocity of the race car.

✡ Formula Used ✡

⭐ v = u + at ⭐

✡ Solution ✡

Given:-

• Initial velocity (u) = 50 m/s
• Acceleration (a) = -10 m/s
• Time taken (t) = 4 sec

▶ According to the question by using formula we get,

$\implies$ v = (50) + (- 10) (4)

$\implies$ v = 50 - 40

$\implies$ v = 10 m/s

Hence, the final velocity of race car is 10m/s.

______________________

⭐ Additional Information ⭐

➡ There are three equation of motion:-

✳ First equation of motion ✴

$\implies$ v = u + at

✳ Second equation of motion ✳

$\implies$ v² = u² + 2as

✳ Third equation of motion ✴

$\implies$ s = ut + $\dfrac{1}{2}$at²

where,

• s = displacement
• u = initial velocity
• v = final velocity
• a = acceleration
• t = time taken

Note :- These equation are reffered as SUVAT equation where SUVAT stands for :-

• s = displacement
• u = initial velocity
• v = final velocity
• a = acceleration
• t = time taken

Explanation:

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