Question

a toy car was moving with a speed of 20cm /s it's speed become 50m/s after traversing a distance of 1m calculate the acceleration of the car​

Answer 1

Given:

Initial Speed of a toy car(u) = 20 cm/s=0.2 m /s

Final Speed (v) =50 m/s

Distance travelled by a toy car=1 m

To Find:

Acceleration of a toy car(a) =?

Solution:

According to the Newton's 3rd equation of motion,

v²= u²-2as

(50) ²=(0.2) ²-2a×1

2500=0.04-2a

2500-0.04= -2a

2499.96= -2a

a= 2499.96/-2

a= -2499. 96/2

a= -1249. 98 m/s²

Here Acceleration is negative acceleration.

Answer 2

According to the question

⇒Initial speed ($u$ ) = 20 cm/s

which can be written as 0.2 m/s

⇒Final speed $( v)$ = 50 m/s

⇒Distance travel = 1m

According to third equation of motion

$v^{2} = u^{2} + 2as$

where,

• v = final speed
• u = initial speed
• a = acceleration
• s = distance travel

⇒ $2as = v^{2} - u^{2}$

⇒ $2a(1) = (50)^{2} - (0.2)^{2}$

⇒ $2a = 2500 - 0.04$

⇒ $2a = 2499.96$

⇒$a = 1249.96 m/ s^{2}$

Hence, the acceleration of the car $a = 1249.96 m/s^{2}$ .

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