{"id":131,"date":"2018-01-15T17:47:48","date_gmt":"2018-01-15T09:47:48","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=131"},"modified":"2018-01-15T17:47:48","modified_gmt":"2018-01-15T09:47:48","slug":"constant-acceleration-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/constant-acceleration-problems-and-solutions.htm","title":{"rendered":"Motion with constant acceleration \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Solved Problems in Linear Motion – Constant acceleration<\/a><\/p>\n

1. A car accelerates from rest to 20 m\/s in 10 seconds. Determine the car’s acceleration!<\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 0 (rest)<\/span><\/span><\/p>\n

Time interval (t) = 10 seconds<\/span><\/span><\/p>\n

Final velocity (v<\/span><\/span>t<\/span><\/sub>) = 20 m\/s<\/span><\/span><\/p>\n

Wanted <\/u><\/span><\/span>: Acceleration (a)<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub> = v<\/span><\/span>o<\/span><\/sub> + a t <\/span><\/span><\/p>\n

20 = 0 + (a)(10)<\/span><\/span><\/p>\n

20 = 10 a<\/span><\/span><\/p>\n

a = 20 \/ 10 <\/span><\/span><\/p>\n

a = 2 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

<\/p>\n

2. A car is decelerating from 30 m\/s to rest in 10 seconds. Determine car’s acceleration.<\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 30 m\/s<\/span><\/span><\/p>\n

Final velocity (v<\/span><\/span>t<\/span><\/sub>) = 0<\/span><\/span><\/p>\n

Time interval (t) = 10 seconds<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> acceleration (a)<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub> = v<\/span><\/span>o<\/span><\/sub> + a t <\/span><\/span><\/p>\n

0 = 30 + (a)(10)<\/span><\/span><\/p>\n

– 30 = 10 a<\/span><\/span><\/p>\n

a = – 30 \/ 10 <\/span><\/span><\/p>\n

a = -3 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

The negative sign appears because the final velocity<\/a> is less than the initial velocity. <\/span><\/span><\/p>\n

<\/p>\n

3. A car starts and accelerates at a constant 4 m\/s<\/span><\/span>2 <\/span><\/span><\/sup>in <\/span><\/span>1 second. Determine speed<\/a> and distance after 10 seconds. <\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

(a) Speed <\/span><\/span><\/p>\n

Acceleration 4 m\/s<\/span><\/span>2<\/span><\/span><\/sup> means speed increase 4 m\/s every 1 second. After 2 seconds, car’s speed is 8 m\/s. After 10 seconds, car’s speed is 40 m\/s. <\/span><\/span><\/p>\n

(b) Distance<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 0<\/span><\/span><\/p>\n

Final velocity (v<\/span><\/span>t<\/span><\/sub>) = 40 m\/s<\/span><\/span><\/p>\n

Acceleration (a) = 4 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Wanted :<\/u><\/span><\/span> Distance<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

s = v<\/span><\/span>o<\/span><\/sub> t + \u00bd a t<\/span><\/span>2 <\/span><\/span><\/sup>= 0 + \u00bd (4)(10<\/span><\/span>2<\/span><\/span><\/sup>) = (2)(100) = 200 meters<\/span><\/span><\/p>\n

<\/p>\n

4. A car travels at a constant 10 m\/s, then decelerates at a constant 2 m\/s<\/span><\/span>2 <\/span><\/span><\/sup>until rest. Determine time elapsed and car’s distance<\/a> before rest.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 10 m\/s<\/span><\/span><\/p>\n

Acceleration (a) = -2 m\/s<\/span><\/span>2<\/span><\/span><\/sup> (The negative sign appears because the final velocity is less than the initial velocity)<\/span><\/span><\/p>\n

Final velocity (v<\/span><\/span>t<\/span><\/sub>) = 0 (rest)<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Time interval and distance<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

(a) Time interval (t)<\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub> = v<\/span><\/span>o<\/span><\/sub> + a t <\/span><\/span><\/p>\n

0 = 10 + (-2)(t)<\/span><\/span><\/p>\n

0 = 10 \u2013 2 t<\/span><\/span><\/p>\n

10 = 2 t<\/span><\/span><\/p>\n

t = 10 \/ 2 = 5 seconds<\/span><\/span><\/p>\n

(b) Distance<\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> = v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup> + 2 a s<\/span><\/span><\/p>\n

0 = 10<\/span><\/span>2<\/span><\/span><\/sup> + 2(-2) s<\/span><\/span><\/p>\n

0 = 100 \u2013 4 s<\/span><\/span><\/p>\n

100 = 4 s<\/span><\/span><\/p>\n

s = 100 \/ 4 = 25 meters<\/span><\/span><\/p>\n

<\/p>\n

5. A car travels at 40 m\/s, decelerates at a constant 4 m\/s<\/span><\/span>2 <\/span><\/span><\/sup>until rest. Determine speed and distance after decelerating in 10 seconds!<\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 40 m\/s<\/span><\/span><\/p>\n

Acceleration (a) = -4 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Time interval (t) = 10 seconds<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> final velocity (v<\/span><\/span>t<\/span><\/span><\/sub>) and distance (s) <\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

(a) Final velocity<\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub> = v<\/span><\/span>o<\/span><\/sub> + a t = 40 + (-4)(10) = 40 \u2013 40 = 0 m\/s<\/span><\/span><\/p>\n

0 m\/s means car rest.<\/span><\/span><\/p>\n

(b) Distance<\/span><\/span><\/p>\n

s = v<\/span><\/span>o<\/span><\/sub> t + \u00bd a t<\/span><\/span>2<\/span><\/span><\/sup> = (40)(10) + \u00bd (-4)(10<\/span><\/span>2<\/span><\/span><\/sup>) = 400 + (-2)(100) = 400 \u2013 200 = 200 meters<\/span><\/span><\/p>\n

<\/p>\n

6. Determine distance after 10 seconds!<\/span><\/span><\/p>\n

\"Constant<\/p>\n

Solution<\/span><\/span><\/p>\n

Distance : <\/span><\/span>s = v t = (10-0)(5-0) = (10)(5) = 50 meters<\/span><\/span><\/p>\n

7. Determine distance after 4 seconds!<\/span><\/span><\/p>\n

\"Constant<\/p>\n

Solution<\/span><\/span><\/p>\n

Distance = square area + triangular area<\/span><\/span><\/p>\n

Distance = (8-0)(8-0) + \u00bd (16-8)(8-0) = (8)(8) + \u00bd (8)(8) = 64 + 32 = 96 meters<\/span><\/span><\/p>\n

8. Determine car’s distance after 4 seconds!<\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

\"Constant<\/p>\n

Distance = triangular area = \u00bd (4-0)(8-0) = \u00bd (4)(8) = 16 meters<\/span><\/span><\/p>\n

9<\/span>. <\/span>A car moves at 90 km\/h past a police car that stops by the side of the road. One minute later, the police car chases <\/span>at <\/span>0.8 m\/s<\/span>2<\/span><\/sup>. How far the police car reach<\/span>es <\/span>the car?<\/span><\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

The speed of car (v) = 90 km\/hour = 90,000 meters \/ 3600 seconds = 25 meters\/second<\/span><\/span><\/p>\n

Time interval (t) = 1 minute = 60 seconds<\/span><\/span><\/p>\n

Acceleration of police’s car (a) = 0.8 m\/s2<\/sup><\/span><\/span><\/p>\n

Initial velocity of police’s car (vo<\/sub>) = 0 m\/s<\/span><\/span><\/p>\n

Wanted :<\/u> Distance traveled by police’s car<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

The car moves at a constant velocity. Distance traveled by the car :<\/u><\/span><\/span><\/p>\n

Initial distance :<\/span><\/span><\/p>\n

s = v t = (25)(60) = 1500 meters<\/span><\/span><\/p>\n

Final distance :<\/span><\/span><\/p>\n

s = v t = (25)(t) <\/span><\/span><\/p>\n

Total distance = 1500 + 25 t<\/span><\/span><\/p>\n

Police’s car moving at a constant acceleration. Distance traveled by police’s car :<\/u><\/span><\/span><\/p>\n

s = v<\/span><\/span>o<\/span><\/sub> t + \u00bd a t<\/span><\/span>2 <\/span><\/span><\/sup>= (0)(t) + \u00bd (0.8)(t<\/span><\/span>2<\/span><\/span><\/sup>) = 0 + 0.4 t<\/span><\/span>2 <\/span><\/span><\/sup>= 0.4 t<\/span><\/span>2 <\/span><\/span><\/sup><\/p>\n

When the police’s car reaches the the car, distance traveled by police’s car same as distance traveled by the car.<\/span><\/span><\/p>\n

Distance traveled by car <\/u><\/span><\/span>= <\/u><\/span><\/span>distance traveled by police’s car<\/u><\/span><\/span><\/p>\n

1500 + 25 t = 0.4 t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

0.4 t<\/span><\/span>2 <\/span><\/span><\/sup>\u2013 25 t \u2013 1500 = 0<\/span><\/span><\/p>\n

Use quadratic formula :<\/span><\/span><\/p>\n

\"Constant<\/p>\n

Distance traveled by police’s car :<\/u><\/span><\/span><\/p>\n

s = 0.4 t<\/span><\/span>2 <\/span><\/span><\/sup>= (0.4)(100<\/span><\/span>2<\/span><\/span><\/sup>) = (0.4)(10,000) = 4000 meter<\/span><\/span>s = <\/span><\/span>4 <\/span><\/span>km<\/span><\/span><\/p>\n

10. A <\/span>car <\/span>moves at a constant 24 m\/s <\/span>brakes so that it has a <\/span>constant deceleration <\/span>of 0.952 m\/s<\/span>2<\/span><\/sup>. <\/span>Determine the speed of the car a<\/span>fter a distance of 250 m<\/span>eters<\/span>.<\/span><\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/span><\/p>\n

Initial velocity (vo<\/sub>) = 24 m\/s<\/span><\/span><\/p>\n

Acceleration <\/span><\/span>(a) = – 0.952 m\/s<\/span><\/span>2 <\/span><\/span><\/sup>(<\/span><\/span>negative signed because deceleration<\/span><\/span>)<\/span><\/span><\/p>\n

Distance <\/span><\/span>(<\/span><\/span>d<\/span><\/span>) = 250 meter<\/span><\/span>s<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Car’s speed after <\/span><\/span>250 meter<\/span><\/span>s<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

Known : initial speed <\/span><\/span>(v<\/span><\/span>o<\/span><\/sub>), <\/span><\/span>acceleration<\/span><\/span> (a), <\/span><\/span>distance <\/span><\/span>(<\/span><\/span>d<\/span><\/span>), <\/span><\/span>wanted : final speed <\/span><\/span>(v<\/span><\/span>t<\/span><\/span><\/sub>) <\/span><\/span>so use the equation of <\/span><\/span>v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> = v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup> + 2 a <\/span><\/span>d<\/span><\/span><\/p>\n

v<\/span><\/span>t <\/span><\/sub>= <\/span><\/span>final velocity<\/span><\/span>, v<\/span><\/span>o<\/span><\/sub> = <\/span><\/span>initial velocity<\/span><\/span>, a = <\/span><\/span>acceleration<\/span><\/span>, <\/span><\/span>d<\/span><\/span> = <\/span><\/span>distance<\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> = (24)<\/span><\/span>2<\/span><\/span><\/sup> + (2)(-0.952)(250)<\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> = 576 \u2013 476<\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> = 100<\/span><\/span><\/p>\n

v<\/span><\/span>t <\/span><\/sub>= \u221a100<\/span><\/span><\/p>\n

v<\/span><\/span>t <\/span><\/sub>= 10 m\/s<\/span><\/span><\/p>\n

<\/p>\n

[wpdm_package id=’507′]<\/p>\n

[wpdm_package id=’517′]<\/p>\n

    \n
  1. Distance and displacement<\/a><\/li>\n
  2. Average speed and average velocity<\/a><\/li>\n
  3. Constant velocity<\/a><\/li>\n
  4. Constant acceleration<\/a><\/li>\n
  5. Free fall motion<\/a><\/li>\n
  6. Down motion in free fall<\/a><\/li>\n
  7. Up and down motion in free fall<\/a><\/li>\n<\/ol>\n

    <\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    Solved Problems in Linear Motion – Constant acceleration 1. A car accelerates from rest to 20 m\/s in 10 seconds. Determine the car’s acceleration! Solution Known : Initial velocity (vo) = 0 (rest) Time interval (t) = 10 seconds Final velocity (vt) = 20 m\/s Wanted : Acceleration (a) Solution : vt = vo + … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Motion with constant acceleration \u2013 problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-131","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/131","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=131"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/131\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=131"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=131"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=131"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}