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Paraphin [41]
3 years ago
14

A minivan is tested for acceleration and braking. In the street-start acceleration test, the elapsed time is 8.6 s for a velocit

y increase from 10 km/h to 100 km/h. In the braking test, the distance traveled is 44 m during braking to a stop from 100 km/h. Assume constant values of acceleration and deceleration.
Physics
1 answer:
Varvara68 [4.7K]3 years ago
6 0

The question is incomplete. Here is the complete question:

A minivan is tested for acceleration and braking. In the street-start acceleration test, the elapsed time is 8.6 s for a velocity increase from 10 km/h to 100 km/h. In the braking test, the distance traveled is 44 m during braking to a stop from 100 km/h. Assume constant values of acceleration and deceleration. Determine  

(a) the acceleration during the street-start test,  

(b) the deceleration during the braking test.

Answer:

(a) 37500 km/h²

(b) 113636.36 km/h²

Explanation:

part (a)

Because it is given that we can assume constant acceleration therefore we can use the following equation of motion:

<em>v = u + (a)(t) </em>

where <em>v </em>is final velocity, <em>u </em>is initial velocity, <em>a </em>is acceleration and <em>t </em>is time change

Given in the question:

v = 100km/h

u = 10 km/h

t = 8.6 sec (changing to hours)

t = 0.0024 hours (round off to 4 decimal places)

100 = 10 + (a x 0.0024)

Rearranging the equation to find value of a

a = (100 – 10) / 0.0024

a = 37500 km/h² (Answer)

part (b)

Now we can use the following equation to find deceleration

<em>2(a)(s) = v² – u²</em>

Where a is acceleration, s is distance travelled, v is final velocity and u is initial velocity

Given in the question

s = 44 m

changing to km

s = 0.044 km

v = 0 km/h (because it stops)

u = 100 km/h

2(a)(0.044) = (0)² – (100)²

0.088(a) = 0 - 10000  

a = - 10000/0.088

a = - 113636.36 km/h2  

The negative sign in the answer shows that it is deceleration

Therefore deceleration = 113636.36 km/h² (Answer)

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Answer:

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Explanation:

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b) The features which affect its stability are;

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3 years ago
Two blocks A and B with mA = 2.2 kg and mB = 0.84 kg are connected by a string of negligible mass. They rest on a frictionless h
madreJ [45]

Answer:

(a) a = 1.875 m/s²

(b)T = 1.575 N

(c)T increase

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the blocks on the horizontal surface and the y-axis in the direction perpendicular to it.

Forces acting on the block  A

WA: Weight of of the A block : In vertical direction  downaward (-y)

NA : Normal force of the A block :In vertical direction  upaward (+y)

F= 5.7 N  In in the direction parallel to the movement of the blocks (+x)

T :  tension of the string: In in the direction (-x)

Forces acting on the block  B

WB: Weight of the B block: In vertical direction  downaward (-y)

NB : Normal force of the B block :In vertical direction  upaward (+y)

T :  tension of the string: In in the direction (+x)

Data

mA = 2.2 kg

mB = 0.84 kg

(a) Magnitude of the acceleration  of the blocks

Newton's second law to B block:

∑Fx = m*a

T = (0.84)*a Equation (1)

Newton's second law to A block:

∑Fx = m*a

5.7 - T = (2.2)*a  We replace T of the Equation (1)

5.7 - (0.84)*a = (2.2)*a  

5.7 = (2.2)*a + (0.84)*a  Equation (2)

5.7 =(3.04)*a

a =5.7 / (3.04)

a = 1.875 m/s²

(b)Tension (in N) in the string connecting the two blocks

We replace data in the  Equation (1)

T = (0.84)*a

T = (0.84)*(1.875)

T = 1.575 N

(c) How will the tension in the string be affected if mA is decreased?

We observed Equation (2)

5.7 = (2.2)*a + (0.84)*a

5.7 = (mA)*a + (0.84)*a

5.7 =a*( mA+ 0.84)

if mA decrease , then, the acceleration increase and T  increase

4 0
3 years ago
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The answer is:

Heat from the ankle is transferred to the ice pack.

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Heat will transfer from the ankle to the ice pack.

7 0
3 years ago
Read 2 more answers
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