Question

A train accelerates from 36 km/hr to 54 km/hr in 10 second. find acceleration and the distance travelled by car​

Answer 1

Answer:

⠀⠀

Accleration of the car is 5m/s$^{2}$

and the distance travelled by the car is 125 m.

⠀⠀

Explanation:

⠀⠀

Given:

⠀⠀

• Initial velocity (u) = 36 km/hr

• Final velocity (v) = 54 km/hr

• Time (t) = 10 s

⠀⠀

To find:

⠀⠀

Acceleration and distance travelled by the car.

⠀⠀

Solution:

⠀⠀

We know that,

Acceleration = $\triangle$ v / $\triangle$ t

$\triangle$ v = a ×$\triangle$ t

⠀⠀

Substitute the known values, we get

$\triangle$ v = 54 – 36 = 18 km/hr.

Now, we need to convert it into m/s

$\implies$ $\triangle$ v = $\frac{18 \times 1000}{3600} \: m/s$

= 5 m/s$^{2}$

⠀⠀

$\implies$ a = 5 m/s$^{2}$

________________________________

⠀⠀

S = ut + 1/2at$^{2}$

Now using this formula, we will find distance travelled.

Substitute the known values, we get

S = 10 × 10 + 1/2 × 5 × 10$^{2}$

$\implies$ S = 125 m.

⠀⠀

Thus, distance travelled by the car is 125 m and acceleration of the car is 5 m/s$^{2}$.

Answer 2

Question :-

A train accelerates from 36 km/hr to 54 km/hr in 10 second. find acceleration and the distance travelled by car.

Answer:

⠀⠀

• Accleration = 5m/s²

• Distance = 125m.

⠀⠀

Given:

• u) = 36 km/hr
• v) = 54 km/hr⠀⠀
• t) = 10 s

Here ,

u = Initial velocity

v = Final velocity

t = Time

⠀⠀

To find:

• ⠀What is the acceleration and distance ?

Solution :⠀⠀

Applying the formula :-

A. = ∆v/∆t

∆v = a × ∆t

Putting the values :- ⠀⠀

∆v = 54 – 36 = 18 km/hr.

Converting km /hr into m/s.

∆v = 18 × 1000/3600 m/s

= 5m/s

Therefore A = 5m/s²

Now , using second equation of motion :- ⠀⠀

S = ut + 1/2at²

Here ,

s = Distance

u = Initial velocity

v = Final velocity

t = Time

Again putting values :-)

s = 10 × 10 + 1/2 × 5 × 10²

s = 125 m .

Hence ,

Acceration = 5m/s².

distance = 125 m .

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