Question

a car starting from rest travels along a straight line covers 96m in 8s.what is the acceleration of the car​

Answer 1

Given :

• Initial velocity, u = 0 m/s (as it is starting from rest)

• Time, t = 8 seconds

• Distance, s = 96 m

To find :

• Acceleration, a

According to the question,

By using Newtons second equation of motion,

➞ s = ut + ½ at²

Where,

• s = Distance
• u = Initial velocity
• a = Acceleration
• t = Time

➞ Substituting the values,

➞ 96 = 0 × 8 + ½ × a × 8 × 8

➞ 96 = 0 + a × 4 × 8

➞ 96 = 0 + 32a

➞ 96 - 0 = 32a

➞ 96 = 32a

➞ 96 ÷ 32 = a

➞ 3 = a

So,the acceleration is 3 m/s².

_____________________________

More formulas :

Newton's first equation of motion :

• v = u + at

Newton's second equation of motion :

• s = ut + ½ at²

Newton's third equation of motion :

• v² = u² + 2as

Answer 2

Answer:

$\huge \bf \: Given$

• Distance covered (D) = 96 M
• Time taken (T) = 8 s
• Initial velocity (U) = 0 m/s

$\huge \bf \: To \: find$

Acceleration of car

$\huge \bf \: Solution$

Here we have to apply Newton second law

$\huge \bf s = ut +\frac{1}{2} \times at {}^{2}$

$\sf \: s \: = distance$

$\sf \: u \: = initial \: velocity$

$\sf \: a \: = acceleration$

$\sf \: t \: = time$

$\huge \tt putting \: value$

$\sf 96 = 0 \times 8 + \dfrac{1}{2} \times a \times 8 \times 8$

$\sf \: 96 = 0 + 1 \times a \times 4 \times 8$

$\sf \: 96 = 1 \times a \times 32$

$\sf \: 96 = 32a$

$\sf \: a \: = 96 \div 32$

$\sf \: a = 3$

$\huge \bf \: acceleration \: = 3 \: mps$