Question

A powerful motorcycle can accelerate from rest to 28 m/s in 4 seconds. What is its average acceleration? How far does it travel in that time​

Answer 1

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Make use of the Followinf kinematics equations

• S = Displacement
• U = initical velocity
• V = final velocity
• A = acceleration

T = time V = U+A+V2 = 02+2ass

= 1/2(U+V)+ts

= ut+1/2at2

The values are as follows

• U = 0m/s
• V = 28m/2
• T = 45
• a = ?? s = ??
• V = u+at
• 28 = 0+49
• 49 = 28 a= 7m/s2

∴the average acceleration = 7m/s2

Displacement = v2 = u2+2as

= 28^2 = 0^2+(2×75)

= 784 = 145

s = 784/14

s = 56

∴Distance travelled = 56m

Answer 2

Step-by-step explanation:

GIVEN

• Initial velocity (u) = 0 m/s --(At rest)
• Final velocity (v) 28 m/s
• Time taken (t) = 4 seconds

TO FIND

• Acceleration (a)
• Distance travelled (s)

SOLUTION

• On using 1st equation of motion, we get,

➥ v = u + at

➱ 28 = 0 + a × 4

➱ 4a = 28

➱ a = 28/4

➱ a = 7 m/s²

Now,

• On using 2nd equation of motion, we get,

➥ s = ut + 1/2 at²

➱ s = 0 × 4 + 1/2 × 7 × (4)²

➱ s = 1/2× 7 × 16

➱ s = 7 × 8

➱ s = 56m

• Acceleration (a) = 7 m/s²
• Distance travelled (s) = 56 m