Question

A bus accelerates uniformly from 54km/h to 72km/h in 10 seconds calculate a) acceleration in m/s 2
b)distance covered in metres

Answer 1

Answer:

a)Acceleration of the bus = 0.5 m/s²

b) distance covered = 175 m

Step by step explanations :

given that,

A bus accelerates uniformly from 54km/h to 72km/h in 10 seconds

here,

we have,

initial velocity of the bus = 54 km/h

54 km/h

= 54 × 1000m/3600s

= 15 m/s

also,

final velocity of the bus = 72 km/h

72 km/h

= 72 × 1000 m/3600 s

= 20 m/s

time taken to change velocity

= 10 seconds

now,

we have,

initial velocity(u) = 15 m/s

final velocity(v) = 20 m/s

time taken(t) = 10 s

to find

a) acceleration

by the equation of motion,

v = u + at

where,

a is the acceleration

putting the values,

20 = 15 + a(10)

10a = 20 - 15

10a = 5

a = 5/10

a = 0.5 m/s²

so,

Acceleration of the bus = 0.5 m/s²

________________

b) distance covered :

here,

we have,

initial velocity(u) = 15 m/s

acceleration(a) = 0.5 m/s²

time taken(t) = 10 seconds

by the equation of motion,

S = ut + ½at²

where,

S = distance covered

putting the values,

S = 15(10) + ½(0.5)(10)(10)

S = 150 + 25

S = 175 m

so,

distance covered by the bus during acceleration will be

175 meter

_____________________

Answers:

a)Acceleration of the bus = 0.5 m/s²

b)distance covered = 175 m

Answer 2

Answer:

(a) Acceleration = 0.5 m/s²

(b) Distance covered in meters = 125 m

Explanation:

Refer to the attachment

sorry for the bad handwriting