Question

A body is moving with a uniform acceleration . its velocity after 5s is 25m/s and after 8s is 34m/s . calculate the distance it will cover in 10sec.​

Answer 1

at t=5 sec initial velocity u = 25m/s

at t=8 sec final velocity v =34m/s

the time interval for which the car travelled = 8-5 = t = 3 sec

therefore the constant acceleration a is given by

a= (v-u)/t

a=(34-25)/3

a=3m/s^2

therefore initial velocity at t=0

u1=v-at

u1=34-3(8)

u1=10m/s

the distance it will cover in the tenth seconds is equal to

(the distance covered after 9 secs) - ( the distance covered after 10 secs)

we use the formula s=ut+1/2 at^2

therfore distance covered after 9 secs = 10(9) + 1/2(3)(9)^2=211.5m

similarly distance covered after 10 sec = 10(10) + 1/2(3)(10)^2=250m

therefore distance covered in the tenth second=250-211.5=38.5m

Answer 2

Explanation:

at t=5 sec initial velocity u = 25m/s 

at t=8 sec final velocity v =34m/s

the time interval for which the car travelled = 8-5 = t = 3 sec 

therefore the constant acceleration a is given by

a= (v-u)/t

a=(34-25)/3

a=3m/s^2

therefore initial velocity at t=0 

u1=v-at

u1=34-3(8)

u1=10m/s

the distance it will cover in the tenth seconds is equal to 

(the distance covered after 9 secs) - ( the distance covered after 10 secs)

we use the formula s=ut+1/2 at^2

therfore distance covered after 9 secs = 10(9) + 1/2(3)(9)^2=211.5m

similarly distance covered after 10 sec = 10(10) + 1/2(3)(10)^2=250m

therefore distance covered in the tenth second=250-211.5=38.5m

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