Question

A train initially traveling at 30 m/s decelerates at 4 m/s22.

a. How much time will it take to slow to 5 m/s?

b. What distance will it have traveled when it has slowed to 5 m/s?


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Answer 1

intial velocity, u = 30 m/s

acceleration, a = – 4 m/s²

{ since it decelerates hence acc is negative }

a) Final velocity, v = 5m/s

using first equation of motion

v = u + at

5 = 30 + ( – 4 ) t

5 – 30 = – 4 t

– 25 = – 4 t

t = 25/4 sec

t = 6.25 sec

b) distance travelled in 6.25 sec

s = ut + 1/2at²

s = 30 ( 6.25 ) + 1/2 ( – 4 ) ( 6.25 )

s = 3 ( 62.5 ) – 0.5 × 4 × 6.25

s = 187.5 – 2 × 6.25

s = 187.5 – 12.5

s = 175 m

Answer 2

♡ Heya user ♡

Given:-

• Initial speed, V1 = 30 m/s

• Declaration, a = -4m/s^2

• Final speed, V2 = 5 m/s^2

Solution :-

a) As we know,

a =( V2 - V1) / t

Substitute the values in the above expression

-4 = (5-30)/t

➯ t = 6.25 sec

Hence, the time taken is 6.25 sec.

b)

As we know,

S = V1 t + 1/2 at^2

Substitute the values in the above expression

S = (30*6.25) + 0.5*-4*(6.25)^2

S = 109.40m

Hence, the distance traveled by the train is 109.40 m.

✧Thanks ✧