Question

A car accelerates uniformly from 36 km/h to 72 km/h in 6s. Calculate the acceleration and the distance covered by the car in that time.​

Answer 1

Given :

• Initial velocity, u = 36 km/s = 10 m/s

• Final velocity, v = 72 km/h = 20 m/s

• Time, t = 6 seconds

To find :

• Acceleration, a &
• Distance covered, s

According to the question,

➞ v = u + at

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time

➞ 20 = 10 + a × 6

➞ 20 - 10 = 6a

➞ 10 = 6a

➞ 10 ÷ 6 = a

➞ 1.67 = a

So,the acceleration is 1.67 m/s².

Now,

➞ s = ut + ½ at²

Where,

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

➞ s = 10 × 6 + ½ × 1.67 × 6 × 6

➞ s = 60 + 1.67 × 3 × 6

➞ s = 60 + 30.06

➞ s = 90.06

So,the distance covered by the car is 90.06 meter.

Answer 2

Answer:

Given :-

Initial velocity (U) = 36 km/h = 10 m/s

Final velocity (V) = 72 km/h = 10 m/s

Time taken (T) = 6 sec

To Find :-

❶ Acceleration

❷ Distance covered

Solution :-

❶ Acceleration

V = U + AT

Here,

V = Final velocity

U = Initial velocity

A = Acceleration

T = Time taken

$\sf \: 20 = 10 + a \times 6$

$\sf \: 20 - 10 = 6a$

$\sf \: 10 = 6a$

$\sf a = 10 \div 6$

$\huge \tt\: Acceleration \: = 1.67 {m}^{2}$

❷Now,

Finding distance travelled

S = ut + ½ at²

Here,

S = Distance travelled

U = Initial velocity

T = Time taken

A = Acceleration

$\sf \: s \: = 10 \times 6 + \dfrac{1}{2} \times 1.67 \times {6}^{2}$

$\sf \: s \: = 60 + \dfrac{1}{2} \times 1.67 \times 36$

$\sf \: s \: = 60 + \dfrac{1}{2} \times 60.12$

$\sf s = 60 + 30.06$

➸ Distance travelled = 90.06 M