Question

the distance of a car moving with velocity 15ms-1 in 2s,decelerates at 2ms-1 is equal to???

Answer 1

Explanation:

s=?

u=15m/sec

t=2 sec

v=2m/sec

a=v-u/t

= 2-15/2

= -13/2

= -6.5 m/sec2

by the third equation of motion (v2= u2 +2as)

4=225+2(-6.5)(s)

4=225-12s

-225+4= -12s

-221 = -12s

-221/-12 =s

s = 18.4 m (approx)

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Answer 2

Given:-

☞Initial Velocity of car = 15m/s

☞Time taken = 2s

☞Deceleration = -2m/s²

 

Find:-

☞Distance Covered by Car-?

 

Formula to be used:-

Using 2nd equation of motion,

☞ $\boxed{\bf{s= ut+\dfrac{1}{2} at^2}}$

 

where,

s= distance

u= initial velocity

t= time

a= acceleration

 

Solution:-

☞Put the given values in the formula we get,

$\sf{\rightarrow s= 15 \times 2+\dfrac{1}{2} (-2) \times 2^2 }$

$\sf{\rightarrow s= 30- 4 }$

$\sf{ \rightarrow s = 26m}$

 

Therefore, Distance Covered by Car is 26m.