Physics, asked by sunepla1306, 1 year ago

a car starting from rest acquires a speed 25 m/s in 10 sec.after which it maintains this speed for 10 sec. find(a) the acceleration (b) distance travelled during acceleration (c) total distance travelled?

Answers

Answered by DSamrat
46

u = 0 m/s


v = 25 m/s


t = 10 m/s


a) a = ?


As v = u + a t


so, 25 = 0 + a * 10


so, a = 25/10 = 2.5 m/s^2


b) s = ? (for first 10 s)


As s = ut + 1/2 a t^2


so, s = 0 + 1/2 * 2.5 * 10^2


so. s = 125 m


c) s = ? ( for next 10 s)


s = v * t as a = 0 m/s^2


so, s = 25*10


so, s = 250 m


Hnce, total distnce = 125 + 250


= 375 m

Answered by sureeshravi
2

Answer:

a) acceleration is 2.5 \frac{m}{s^2}

b) distance during acceleration 125 m

c) total distance 375 m

Explanation:

GIVEN:

initial speed u = 0 m/s

final speed  v = 25 m/s

time taken t = 10 s

Determining a) Acceleration a in \frac{m}{s^2}

acceleration is given by equation as given below:

a = \frac{changeInVelocity}{changeInTime}

a = \frac{v-u}{t}\\

substituting value in above equation

a = \frac{25-0}{10}\\ a = 2.5 \frac{m}{s^2}

Determining b) distance travelled during acceleration, lets say it be s

s = ut + \frac{at^2}{2}\\

Substituting value in equation

s = 0*10 + 0.5*2.5*(10^2)\\s = 0 + 1.25*100\\s = 125 m

Determining c) total distance travelled, be d

d = s + y

Lets say y is distance travelled through uniform motion

y = speed * time\\y = 25*10\\y = 250 m

Now adding them together gives d

d = 125 + 250\\d = 375m

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