Question

A bus is travelling at speed of 15m/s. Then it accelerates for 5s at a rate of 10m/s2. Find the final velocity and distance covered by the bus during this process.​

Answer 1

Answer:

Step-by-step explanation:

Initial velocity = u = 15 ms-¹

Acceleration = a = 10 ms-²

Time taken = t = 5s

Now,

Using first equation of motion we can find the final velocity,

v = u + at

v = 15 + 10 × 5

v = 65 ms-¹

Again,

we can solve the distance covered by using both second and third equation of motion.

So using second equation,

S = ut + 1/2 at²

S = 15 × 5 + 1/2 × 10 × 5 × 5

S = 75 + 125

S = 200 m

Using third equation,

v² - u² = 2aS

(v-u)(v+u) = 2(10)S [v² - u² = (v-u)(v+u)

(65-15)(65+15) = 20S

20S = 4000

S = 200 m

Hence,

The final velocity = v = 65 ms-¹

Distance covered = S = 200 m

Answer 2

Given :-

• The speed of the bus = 15m/s
• The acceleration of bus = 10m/s²
• Time taken by the bus = 5s

To Find :-

• The final velocity of bus (v) = ?
• The distance covered by the bus (s) = ?

Solution :-

• To calculate the final velocity of bus and distance travelled by the bus , at first we have to use first equation of motion to calculate final velocity of the bus then calculate distance travelled by the bus by applying 2ns equation of motion.

Calculation for Final velocity :-

• V = ? U = 15m/s. T = 5s. A = 10m/s² :-

⇢ First equation of motion :-

⇢ V = U + A × T

⇢ V = 15 + 10 × 5

⇢ V = 15 + 50

⇢ V = 65m/s

Therefore, final velocity of the bus (v) = 65m/s

Calculation for Distance covered by bus :-

• S = ? U = 15m/s. T= 5s. A = 10m/s²

⇢ 2nd equation of motion :-

⇢ S = UT + 1/2 × A × T²

⇢ S = 15 × 5 + 1/2 × 10 × 5²

⇢ S = 75 + 5 × 25

⇢ S = 75 + 125

⇢ S = 200m

Therefore, distance covered by the bus (s) = 200m