A car initially at rest starts moving with a constant acceleration of 0.5 m/s2 and travels a distance of 25 m. Find 5m (i) Its final velocity. (ii) Time taken
Answers
Required Answer :
- The final velocity of the car = 5 m/s
- The time taken by the car = 10 seconds
Given :
- Initial velocity of the car = 0 m/s
- Acceleration of the car = 0.5 m/s²
- Total distance travelled by the car = 25 m
To find :
- Final velocity of the car
- Time taken by the car
Solution :
To calculate the final velocity of the car, we will use the third equation of motion.
Third equation of motion :
- v² - u² = 2as
where,
- v denotes the final velocity
- u denotes the initial velocity
- a denotes the acceleration
- s denotes the distance travelled/displacement
Substituting the given values :
⇒ v² - (0)² = 2(0.5)(25)
⇒ v² - 0 = 2 × 5/10 × 25
⇒ v² = 5/5 × 25
⇒ v² = 25
⇒ Taking square root on both the sides :
⇒ v = √25
⇒ v = √(5 × 5)
⇒ v = 5
Therefore, the final velocity of the car = 5 m/s
To calculate the time taken by the car, we will use the first equation of motion.
First equation of motion :
- v = u + at
where,
- v denotes the final velocity
- u denotes the initial velocity
- a denotes the acceleration
- t denotes the time taken
Substituting the given values :
⇒ 5 = 0 + 0.5 × t
⇒ 5 = 0.5 × t
⇒ 5/0.5 = t
⇒ 5 × 10/5
⇒ 10 = t
Therefore, the time taken by the car = 10 seconds
Given Initial velocity u = 0
Acceleration a = \rm 0.5m {s}^{ - 2}0.5ms
−2
Distance travelled S = 25m
i)Final velocity
From equation of motion \rm {v}^{2}={u}^{2}+2aSv
2=u
2+2aS
\implies \rm \: {v}^{2} = {0}^{2} + 2 \times 0.5 \times 25⟹v
2=0
2 +2×0.5×25
implies {v}^{2} = 25⟹v
2=25
implies {v} =sqrt{25}⟹v=
25
\rm \fbox{ \therefore \: final \: velocity \: v \: = 5 ms{}^{ - 1} }
ii) Time taken
From equation of motion v = u + at
implies⟹ 5 = 0 + 0.5 × t
implies⟹ 0.5 × t = 5
{t = \frac{5}{0.5} }⟹t= 5/0.5
therefore time taken = 10s}