Question

A race car traveling on a straight part of a track has an initial velocity of 10.0 m/s. If the car then accelerates at 2.00 m/s2, how far will the car travel in 3.0 seconds?

Answer 1

Answer :

➥ The Distance travelled by a race car = 39 m

Given :

➤ Initial velocity of a race car = 10.0 m/s

➤ Acceleration of a race car = 2.00 m/s²

➤ Time taken by a car = 3.0 sec

To Find :

➤ Distance travelled by a race car = ?

Required Solution :

For solve this question, let's first know about Distance.

Distance is defined as how much ground an object has covered.

• Distance is a scalar quantity.
• The SI unit of Distance is m.
• The unit of Distance is cm in the CGS system.
• Distance in represented by s.

So let's solve this question...

❒ According to the given Question, given that Initial velocity is 10.0 m/s so, we can write 10 m/s, Acceleration is 2.00 m/s² so, we can write 2 m/s², Time taken is 3.0 sec so we can write 3 sec. Because zero has no value after the point.

Now, we have Initial velocity, Acceleration, and it's time taken by a race car,

• Initial velocity of a race car = 10 m/s
• Acceleration of a race car = 2 m/s²
• Time taken by a race car = 3 sec

We can find Distance travelled by a race car by using the second equation of motion which says s = ut + ½ at².

Here,

• s is the distance in m.
• u is the Initial velocity in m/s.
• t is the Time taken in second.
• a is the Acceleration in m/s².

✎ So let's find Distance travelled (s) !

From second equation of motion

⇛ s = ut + ½ at²

⇛ s = 10 × 3 + ½ × 2 × 3²

⇛ s = 10 × 3 + ½ × 2 × 9

⇛ s = 30 + ½ × 2 × 9

⇛ s = 30 + 1 × 9

⇛ s = 30 + 9

⇛ s = 30 m

║Hence, the distance travelled by a race car is 30 m.║

$\:$

Some related equations :

⪼ First equation of motion: v = u + at

⪼ Second equation of motion: s = ut + ½ at²

⪼ Third equation of motion: v² = u² + 2as

Where,

• v is the Final velocity in m/s.
• u is the Initial velocity in m/s.
• a is the Acceleration in m/s².
• t is the Time taken second.
• s is the Distance in m.

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Answer 2

Answer:

$\tt {\pink{We \:have}}\begin{cases} \sf{\green{Initial \: Velocity (u)= 10 \: m/s}}\\ \sf{\blue{Acceleration \: (a) = 2 \: m/s^2 }}\\ \sf{\orange{Time \: taken (t)=3 \: seconds}}\\ \sf{\red{Distance \: (s)= \: ?}}\end{cases}$

______________________

$\: \: \: \: \qquad\tiny\dag \: \underline{\textsf{ \textbf{ By using third equation of motion : }}}$

$:\implies \sf s = ut + \dfrac{1}{2} \: at^2$

$:\implies \sf s = 10 \times 3 + \dfrac{1}{\cancel{2}} \times \cancel{2} \times {3}^{2}$

$:\implies \sf s = 30 + 9$

$:\implies \textsf{ \textbf{s = 39 \: m}}$