Question

A car accelerate uniformly from 13 km/hr for to 36 km hr in
5 sec Calculate i] The acceleration ii] The distance covered
in that time
by the
car​

Answer 1

Solution :—

Given :-

• Initial Velocity (u) = 13 km/hr = 3.62 m/s
• Final Velocity (v) = 36 km/hr = 10 m/s
• Time Taken (t) = 5 s

To Find :-

• Acceleration (a)
• Distance Covered (s)

Calculating Acceleration

Formula Used :-

• v = u + at

⟹ 10 = 3.62 + a × 5

⟹ 10 = 3.62 + 5a

⟹ 10 - 3.62 = 5a

⟹ 6.38/5

⟹ 1.276

Thus, acceleration is 1.276 m/s²

Calculating Distance Covered

Formula Used :-

• v² - u² = 2as

⟹ (10)² - (1.276)² = 2 × 6.38 × s

⟹ 100 - 1.628176 = 12.76 s

⟹ 98.371824 = 12.76 s

⟹ s = 86.8956/12.76

⟹ s = 7.71

Thus, Distance Covered is 7.71 m

Hence, Solved ♡

Answer 2

Question

A car accelerates uniformly from 13 km/hr to 36 km/hr in 5 sec. Calculate -

i) Acceleration

ii) Distance travelled by car.

Answer

Given -

• u = 13 km/hr
• v = 36 km/hr
• t = 5 sec

To find -

Acceleration - a

Distance travelled - s

Formula used -

$\boxed{\rm \gray {a = \frac{v-u}{t}}}$

$\boxed{\rm \gray {s = ut + {}_{1}/{}_{2} at^2}}$

Solution -

Finding Acceleration -

• u = 13 km/hr = 3.6 m/s
• v = 36 km/hr = 10 m/s
• t = 5 sec

Substituting value in formula -

$\rm a = \frac{v-u}{t}$

$\rm a = \frac{10-3.6}{5}$

$\rm a = \frac{6.4}{5}$

$\rm a = 1.28 m/s^2$

Acceleration = 1.28 m/s²

Finding Distance Travelled -

• u = 13 km/hr = 3.6 m/s
• a = 1.28 m/s²
• t = 5 sec

Substituting the value in 2nd equation of motion -

$\rm s = ut + 1/2 at^2$

$\rm s = 3.6 \times 5 + 1/2 \times 1.28 \times 25$

$\rm s = 18 + 16$

$\rm s = 34 m$

Distance travelled = 34 m