Question

Q#2: The velocity of a Bus is reduced uniformly from 15 m/s to 7 m/s while

traveling a distance of 90 m
1. Compute the acceleration,
How much further will the Bus travel before coming to rest, provided the acceleration
remain constant?​

Answer 1

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Answer 2

Answer:-

• Acceleration of the bus = -0.98m/s²

• Distance travelled by the bus further

before coming to rest = 24.8m

Explanation:-

Firstly, let's calculate the acceleration of the bus by using the 3rd equation of motion :-

=> v² - u² = 2as

Where:-

• v is final velocity of the bus .

• u is initial velocity of the bus.

• a is acceleration of the bus.

• s is distance travelled.

=> (7)² - (15)² = 2×a×90

=> 49 - 225 = 180a

=> -176 = 180a

=> a = -176/180

=> a = -0.98m/s²

Now in the 2nd case, we have to calculate the distance covered by the bus further before coming to rest.

Hence, in this case :-

• Final velocity of the bus will be zero as

it finally comes to rest.

=> v² - u² = 2as

=> 0 - (15)² = 2(-0.98)s

=> -225 = -1.96s

=> s = -225/-1.96

=> s = 114.8m

So, the bus travels a total distance of 114.8m before coming to rest. But we have to calculate the distance travelled by the car after it had initially covered 90m.

Thus, distance travelled furtherwards :-

= 114.8 - 90

= 24.8m