Question

A skateboarder, starting from rest, rolls down a 10.9-m ramp. When she arrives at the bottom of the ramp her speed is 6.74 m/s. (a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 26.6 ° with respect to the ground, what is the component of her acceleration that is parallel to the ground

Answer 1

Answer:b

Explanation:b

Answer 2

The magnitude and the component of her acceleration will be:

• (a) 2.08 m/s²
• (b) 1.85 m/s²

According to the question,

Side of ramp,

• s = 10.9 m

Speed,

• v = 6.74 m/s

Angle,

• θ = 26.6°

(a)

As we know,

→ $v^2=u^2+2as$

or,

→ $a = \frac{v^2}{2s}$

By substituting the values, we get

→    $= \frac{(6.74)^2}{2\times 10.9}$

→    $= \frac{45.42}{21.8}$

→    $= 2.08 \ m/s^2$

(b)

→ $a \ CosA$ = $2.08\times Cos 26.6^{\circ}$

→              = $2.08\times 0.89$

→              = $1.85 \ m/s^2$

Thus the above answer is correct.

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