# A car accelerates uniformly from 18 km/h to 36 km/h in 5 seconds.calculate the -a)accelerationb)distance covered by the car

gdgdgdgdgdg

time is given 5 second the right solution will be

u = 18*5/18=5

v= 36*5/18=10

acceleration= v-u/t = 10-5/5=5/5=1m/sec^2

distance = s

So, s = ut + 1/2 at2

= 5 * 5 +1/2 * 1 (5)2

= 25 + 25 / 2

= 50+25/2

= 75/2

= 37.5 m

dont thumb down any answer if u dont know how to doo

• 67

u = 18*5/18=5

v= 36*5/18=10

acceleration= v-u/t   = 10-5/5=5/5=1m/sec^2

• -1

here it is:

18 km / hr = 18 * 1000/ 3600 = 5m/sec.

36 km/hr = 36 * 1000 / 3600 = 10m/sec.

so,accelaration = v-u/t

=10 - 5 /5

=1 m/ sec.

distance = s

So, s = ut + 1/2 at2

= 5 * 5 +1/2 * 1 (5)2

= 25 + 25 / 2

= 50+25/2

= 75/2

= 37.5 m

• 27

here,

u = 18 km/hr = 18 x 5/18 m/s = 5 m/s

v = 36 km/hr = 10 m/s

so, acceleration a = (v - u) / t = (10-5) / 10 = 0.5 m/s2

now, the total distance travelled will be

s = ut + (1/2)at2

or

s = 5x10 + (1/2).0.5.(10)2

or

s = 50 + 25

thus,

s = 75m

• -2

gdgdgdgdgdg

time is given 5 second the right solution will be

u = 18*5/18=5

v= 36*5/18=10

acceleration= v-u/t = 10-5/5=5/5=1m/sec^2

s=ut + (1/2)at2

or

s = 5x10 + (1/2).0.5.(10)2

or

s = 50 + 25

thus,

s = 75m

• 1

gdgdgdgdgdg

time is given 5 second the right solution will be

u = 18*5/18=5

v= 36*5/18=10

acceleration= v-u/t = 10-5/5=5/5=1m/sec^2

s=ut + (1/2)at2

or

s = 5x10 + (1/2).0.5.(10)2

or

s = 50 + 25

thus,

s = 75m

this will be right

• 3
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