Question

A bus starts from rest and attains a velocity of 46m/s after 10sec.Find the acceleration and the distance travelled in 5sec and in 10sec.​

Answer 1

Answer:-

Given:

Initial Velocity (u) = 0 m/s [ bus is at rest ]

Final Velocity (v) = 46 m/s

Time (t) = 10 sec

We know that,

Acceleration (a) = (v - u) / t

→ a = (46 - 0) / 10

→ a = 46/10

→ a = 4.6 m/s²

We have to find:

The distance travelled in 5 s and 10 s.

Using 2nd equation of motion,

→ s = ut + 1/2 * at²

[ s is the Distance travelled ]

The value of s when t = 5 s is :

→ s = (0)(5) + 1/2 * (4.6) * (5)²

→ s = 57.5 m

The value of s when t = 10 s is :

→ s = (0)(10) + 1/2 * (4.6) * (10)²

→ s = 230 m

Therefore,

• Acceleration of the bus is 4.6 m/s²
• Distance travelled in 5 seconds is 57.5 m.
• Distance travelled in 10 seconds is 230 m.

Answer 2

Answer:

Explanation:

OK the question is quite easier.

The acceleration is the rate of change of velocity that needs to be understood first.

a ==> acceleration

u==> initial velocity

v==> final velocity

s==> distance travelled

t==> time

Over here as the bus is starting from rest, the initial velocity = u = 0 m/s

final velocity attained = v = 46 m/s

we have the formula.

a = [v-u] / t

= 46-0 / 10

= 46/10 = 4.6 m/s²

thus we say a =4.6 m/s²

at t=5

v = at = 4.6 x 5 = 23.0 m/s

v' = 46 m/s

s = ut + 1/2 at²

= (23)(5) + 1/2 (4.6)(5)²

172.5 m

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