# A car accelerates uniformly from rest to a speed of 24.3 km/h in 5.8 s. Find the distance it travels during this time in meters.

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### 2 Answers

A car accelerates uniformly from rest to a speed of 24.3 km/h in 5.8 s.

The initial velocity of the car is u = 0 and the final velocity is v = 24.3 km/h = 6.75 m/s

The distance traveled by the car in this duration is `u*t + (1/2)*a*t^2`

=> `0*5.8 + (1/2)*(6.75 - 0)/5.8*5.8^2`

=> `0 + 6.75*5.8/2`

=> 19.575 m

**The car travels a distance 19.575 m.**

Because the car accelerates uniformly from the rest (v_0=0 m/s, s_0=0m) to 24.3 km/h (v_f),

hence, we can apply the formular a=(v_f-v_i)/t, and convert the known v_f and t into the same uits.

24.3 km/h=24.3*1000m/(1h*3600s/h)= 6.75 m/s.

a=(v_t-v_i)/t=(6.75m/s-0m/s)/5.8s=1.16 m/s^2.

By using s_total=s_0+V_i*t+1/2*a*t^2=0 m+0 m/s^2*5.8+1/2*1.16 m/s^2*(5.8 s)^2=0 m+19.5 m=19.5m.