Question

Starting from rest a car acquires a velocity of 30km/h in 10 s and then brakes are applied ,it take 20s to stop.Calculate the distance travelled by the car. *​

Answer 1

Answer:

Distance travelled by the car= 83m

Explanation:

Initial velocity= 0m/s (as the car starts from rest)

Now,

Let us find acceleration in 10s

so, u = 0m/s

v = 30km/h=8.3m/s

t = 10s

Acceleration= v-u/t

= 8.3-0/10= 0.83m/s2

Now distance in 10s=

v2-u2=2as

(8.3)2-0=2*0.83s

68.89=1.66s

s= 68.89/1.66

s=41.5m

Now after that,brakes are applied and it took 20s to stop, so there is decelerating

Therefore, acceleration in 20s = -0.83m/s2( minus sign indicates that there is deceleration)

Now let us calculate the distance in 20s

here,

Initial velocity = 30km/h=8.3m/s

Final velocity= 0m/s(as it comes to rest)

Time=20s

v2-u2=2as

0- (8.3)2=2*(-0.83)*s

-68.89=1.66s

s=68.89/1.66

s=41.5m

So Total Distance covered = (41.5+41.5)m

= 83m

Answer 2

During acceleration,

v = u + at

30×5/18 = 0 + a(10)

a = 5/6 m/s²

S(1) = ut + 1/2 at²

= 0 + 1/2 (5/6) (10)²

= 250/6 = 41.66 m

During deceleration,

v = u + at

0 = 30×5/18 + a(20)

a = -5/12 m/s²

S(2) = ut + 1/2 at²

= (30×5/18)(20) + 1/2 (-5/12)(20)²

= 500/3 + 500/6

= 1500/6 = 250m

Total distance travelled

= S(1) + S(2) = 291.66m