Question

5m/s^2 retardation is given for a body moving with a velocity 40 m/s.
a) Find the time taken by the body to come to a stop
b) What is the displacement in this time?

Answer 1

a) time = 8s b) displacement = 160m

Given,

Retardation a = -5m/s^2

Initial velocity u = 40m/s

Final velocity v = 0m/s {since body comes to rest}

a) Using v = u + at

0 = 40 - 5t

t = 8

b) Using (v^2 - u^2)/2a = s

-1600/-10 = s

s = 160m

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Answer 2

Given :

• Final velocity (v) = 0 m/s
• Initial velocity (u) = 40 m/s
• Acceleration (a) = - 5 m/s²

Negative sign means retardation

To find :

• Time taken to stop
• Displacement (s)

According to the question,

i)

By using Newtons first equation of motion,

→ v = u + at

Where,

• v = Final velocity
• u = Initial velocity
• t = Time
• a = Acceleration

→ Substituting the values,

→ 0 = 40 + (-5) × t

→ 0 - 40 = -5t

→ -40 = -5t

→ 40 ÷ 5 = t

→ 8 = t

So,the time taken by the body to stop is 8 seconds.

ii)

By using Newtons second equation of motion,

→ s = ut + ½ at²

Where,

• s = Distance
• u = Initial velocity
• a = Acceleration
• t = Time

→ Substituting the value,

→ s = 40 × 8 + ½ × (-5) × 8 × 8

→ s = 320 + (-5) × 4 × 8

→ s = 320 - 160

→ s = 160

So,the displacement is 160 metres.