Question

A car starts from rest and reaches a velocity of 50 m/s. If it took 10 seconds to reach that velocity,
what is the distance covered in those 10 seconds ?
250 m
200 m
160 m
ОО
166.67 m

Answer 1

AnsWer :

Given;

•Initial velocity (u) = 0 m/s

•Final velocity (v) = 50 m/s

•Time Interval (t) = 10 seconds.

Unknown Quantities;

•Acceleration (a) = ?

•Distance (s) = ?

Solution;

● A C C E L E R A T I O N :

•By applying first kinematical equation of motion we have:

↠v = u + at

↠v - u = at

↠a = (v - u)/t

↠a = (50 - 0)/10

↠a = 50/10

↠a = 5 m/s²

● D I S T A N C E :

•By applying third kinematical equation of motion we have:

⇥v² - u² = 2as

⇥(50)² - (0)² = 2 × 5 × s

⇥2500 = 10 × s

⇥s = 2500 ÷ 10

⇥s = 250 m

∴The distance covered in those 10 seconds is option A) 250 m.

Answer 2

Initial velicity of the car, u = 0 m/s

Final velocity of the car, v = 50 m/s

Time taken to reach the final velocity, t = 10 s

Distance covered by the car = s

By using the formula, $\sf v = u + a·t$

• $\sf 50 = 0 + a·10$

• $\sf a = 5 m·s^2$

Now, as we know, $\sf s = u·t + \dfrac{1}{2}·a·t^2$

• $\sf s = 0·10 + \dfrac{1}{2}.5·[10]^2$

• $\sf s = \dfrac{1}{2}·5·100$

• $\sf s = 5·50$

• $\sf s = 250 m$

Therefore, the distance covered in those 10 seconds is 250 m.