Question

A sport car starts from rest and accelerates with constant acceleration of 5 m/s2 for 13 seconds.
(a) What is the final velocity of the sport car?
(b) What is the total distance travelled?

Answer 1

Step-by-step explanation:

Given:-

• Initial velocity (u) = 0m/s

• Acceleration (a) = 5m/s²

• Time (t) = 13 seconds

To Find:-

• The final velocity of the car.

• The total distance travelled by the car.

Solution:-

• u = 0m/s , $\: \:$ a = 5m/s² , $\:$ t = 13 s, $\:$ v = ? and s = ?

Using 1st equation of motion:-

$\leadsto \underline{ \: \: \underline{ \: \textbf{ \pink{v = u + at}} \: } \: \: }$

$\dashrightarrow \sf v = 0 + (5 \times 13)$

$\dashrightarrow \sf v = 65$

$\underline{ \boxed{ \therefore\textsf{\textbf{ \red{ Final \: velocity = 65m/s}} \: } }}$

Using 3rd equation of motion:-

$\leadsto \underline{ \: \: \underline{ \: \bf{ \green{s= ut + \dfrac{1}{2}a {t}^{2} }} \: } \: \: }$

$\dashrightarrow \sf s =( 0 \times 13) + \dfrac{1}{2} \times 5 \times {(13)}^{2}$

$\dashrightarrow \sf s = \dfrac{1}{2} \times 5 \times 169$

$\dashrightarrow \sf s = \dfrac{845}{2}$

$\dashrightarrow \sf s = 422.5m$

$\underline{ \boxed{ \therefore\textsf{\textbf{ \purple{ Distance \: travelled = 422.5m}} \: } }}$

Answer 2

Answer:

• The final velocity of the sport car is 65 m/s.

• The total distance travelled is 422.5 m.

Explanation:

Given that,

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 5 m/s²
• Time (t) = 13 sec

As we know that,

$\purple\bigstar$ a = (v - u)/t

[ Putting values ]

↪ 5 = (v - 0)/13

↪ 5 × 13 = v

↪ v = 65 m/s $\red\bigstar$

Applying 2nd eqn. of motion,

$\green\bigstar$ s = ut + 1/2 × at²

[ Putting values ]

↪ s = 0 × 13 + 1/2 × 5 × (13)²

↪ s = 2.5 × 169

↪ s = 422.5 m $\red\bigstar$

For information:

• v = u + at
• v² = u² + 2as