Question

10. A car accelerating uniformly from 15m/sec to 20 m/sec in 8 sec calculate
a) Its acceleration b) The distance covered by the car in that time.​

Answer 1

✬ Acceleration = 0.6 m/s² ✬

✬ Distance = 139.2 m ✬

Explanation:

Given:

• Uniform velocity (u) of car is 15 m/sec.
• Final Velocity (v) of car is 20 m/s.
• Time (t) taken by car is 8 seconds.

To Find:

• What is the acceleration of car and distance covered by it ?

Formula to be used:

• v = u + at ( For acceleration )
• s = ut + 1/2at² ( For distance )

Solution: Using the first equation of motion

$\implies{\rm }$ v = u + at

$\implies{\rm }$ 20 = 15 + a $\times$ 8

$\implies{\rm }$ 20 – 15 = 8a

$\implies{\rm }$ 5 = 8a

$\implies{\rm }$ 5/8 = a

$\implies{\rm }$ 0.6 m/s² = a

Hence, the acceleration of the car is 0.6 m/s².

Now, For finding distance use the second equation of motion

$\implies{\rm }$ s = ut + 1/2at²

$\implies{\rm }$ s = 15 $\times$ 8 + 1/2 $\times$ 0.6 $\times$ 8²

$\implies{\rm }$ s = 120 + 0.6/2 $\times$ 64

$\implies{\rm }$ s = 120 + 0.3 $\times$ 64

$\implies{\rm }$ s = 120 + 19.2

$\implies{\rm }$ s = 139.2 m

Hence, the distance covered by the car is 139.2 m.

Answer 2

Explanation:

Acceleration = 0.6 m/s² ✬

✬ Distance = 139.2 m ✬

Explanation:

Given:

Uniform velocity (u) of car is 15 m/sec.

Final Velocity (v) of car is 20 m/s.

Time (t) taken by car is 8 seconds.

To Find:

What is the acceleration of car and distance covered by it ?

Formula to be used:

v = u + at ( For acceleration )

s = ut + 1/2at² ( For distance )

Solution: Using the first equation of motion

v = u + at

20 = 15 + a 8

20 – 15 = 8a

5 = 8a

5/8 = a

0.6 m/s² = a

Hence, the acceleration of the car is 0.6 m/s².

Now, For finding distance use the second equation of motion

s = ut + 1/2at²

s = 15 8 + 1/2 0.6 8²

s = 120 + 0.6/2 64

s = 120 + 0.3 64

s = 120 + 19.2

s = 139.2 m

Hence, the distance covered by the car is 139.2 m