Question

A bus accelerates uniformly from 72 m/s 200 m/s in 20 seconds calculate acceleration and the distance travelled by the first during this interval of time

Answer 1

Explanation:

hello there...

this answer to this question is..

acceleration is = (v-u)/t

where v= final velocity

u= initial velocity

thus acceleration = 200-72/20

= 6.4 m/s^2

hope you understood

have a great day ahead!!

Answer 2

Answer:-

$\red{\bigstar}$ Acceleration = 6.4 m/s²

$\red{\bigstar}$ Distance travelled = 2720 m.

• Given:-

Initial velocity [u] = 72 m/s

Final velocity [v] = 200 m/s

Time [t] = 20 sec.

• To Find:-

Acceleration [a] = ?

Distance travelled [s] = ?

• Solution:-

• Using 1st equation of motion:-

$\pink{\bigstar}$ $\boxed{\sf{v = u + at}}$

→ 200 = 72 + a × 20

→ 200 - 72 = 20 × a

→ 20 a = 128

→ a = $\dfrac{128}{20}$

→ a = 6.4 m/s²

Now,

• Using 2nd equation of motion:-

$\pink{\bigstar}$$\boxed{\sf{s= ut + \dfrac{1}{2} a t^{2}}}$

→ s = 72 × 20 + $\dfrac{1}{2}$× 6.4 × (20)²

→ s = 1440 + 3.2 × 400

→ s = 1440 + 1280

→ s = 2720 m

Therefore, the acceleration is 6.4 m/s² and distance travelled is 2720 m.