Rohangrr 3 years ago okay @lgbasallote @Shane_B @baddinlol A racing car has a uniform acc.. of $4 \quad m/s^2$. What's the distance she will cover 10s later

Does it start from rest?

2. Rohangrr

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3. Rohangrr

thanks

Wait

Its wrong

Use the formula v = ut + 0.5at^2

7. Shane_B

Assuming a start from rest: $d=Vi(t)+\frac{1}{2}at^2=0m/s^2+\frac{1}{2}(4m/s^2)(10s)^2=200m$

Oops sorry

9. Rohangrr

lol yes firstly u found the mistake

x = ut + 0.5at^2

11. Rohangrr

@Shane_B can u make it more clearer

12. Shane_B

Or, you can use: $d=\frac{V_i+V_f}{2}t=\frac{0m/s+40m/s}{2}(10s)=200m$

13. Shane_B

Which part isn't clear?

14. Rohangrr

well plzzz make it clear from the beggining

Well he used this formula x = ut + 0.5at^2

u = 0m/s , becausehe assumed the car started from rest

t = 10s

a = 4m/s^2

Then he plugged in the numbers x = 0.5*4* 100 = 200m

20. Shane_B

@baddinlol: thanks...working with someone else at the moment :)

21. Rohangrr

@Shane_B ITS $\huge 4 \quad m/s^2$

22. Rohangrr

not 40 m/s2

no no

The final velocity is 40m/s

25. Shane_B

40m/s is the final velocity...

26. Rohangrr

okay lol